Quadratic Functions — Free Algebra 2 Review Games.

Vertex form is y = a(x-h)^2 + k where (h,k) is the vertex.

In vertex form y = (x-h)^2+k, vertex is (h,k) = (3,2).

Negative leading coefficient means the parabola opens downward.

Axis \(= -b/(2a) = 6/2 = 3\).

By definition, i is the imaginary unit where i^2 = -1.

Take half of 8 (which is 4) and square it: 16.

D = b^2-4ac = 9-40 = -31.

A negative discriminant means two complex conjugate roots.

\(\sqrt{-16} = \sqrt{16}\sqrt{-1} = 4i\).

Complete the square: \(x^2-4x+4+3 = (x-2)^2+3\).

\(x^2 = -4\), \(x = \pm 2i\).

3-3i+2i-2i^2 = 3-i+2 = 5-i.

x = -12/(2*2) = -3. y = 2(9)+12(-3)+7 = 18-36+7 = -11.

Substitute \(x=2\): \(4-12+k=0\), \(k=8\).

Multiply by conjugate: (2+3i)(1+i)/((1-i)(1+i)) = (2+2i+3i+3i^2)/(1+1) = (-1+5i)/2.

Using the cycle of powers of i: i^1 = i, i^2 = -1, i^3 = i^2 · i = (-1)(i) = -i. A common mistake is stopping at i^2 and choosing -1, but the exponent is 3, not 2.

Substitute x = 0: y = (0 - 2)^2 + 5 = 4 + 5 = 9. A common error is reading off the vertex value k = 5 as the y-intercept, but the y-intercept requires setting x = 0, not reading the vertex.

In vertex form y = a(x - h)^2 + k, the axis of symmetry is x = h. Rewriting: y = (x - (-3))^2 - 7, so h = -3 and the axis is x = -3. The distractor x = 3 is a sign error caused by ignoring that (x + 3) means h = -3.

When b^2 - 4ac = 0, the quadratic formula gives x = -b / (2a), which is a single value repeated. This is called a double root. Two distinct real solutions require discriminant > 0, and no real solutions occur when discriminant < 0.

A parabola opens upward when the leading coefficient \(a\) is positive. In \(y = 2(x - 1)^2 + 3\), \(a = 2 > 0\), so it opens upward. All three other choices have a negative leading coefficient, causing them to open downward.

In vertex form y = a(x - h)^2 + k, the vertex is (h, k) and k is the minimum value when a > 0. Here a = 1 > 0 and k = -4, so the minimum value is -4. The distractor 4 ignores the negative sign, and -1 confuses h with k.

i^4 = (i^2)^2 = (-1)^2 = 1. Equivalently, the powers of i cycle every 4 steps: i, -1, -i, 1. The fourth step always returns to 1. The distractor -1 is the value of i^2, not i^4.

To complete the square for \(x^2 + bx\), add \((b/2)^2\). Here \(b = -10\), so \((-10/2)^2 = (-5)^2 = 25\). The distractor 100 is \(b^2 = (-10)^2\), a common error of squaring \(b\) without first dividing by 2.

Complete the square: \(x^2 + 6x = (x + 3)^2 - 9\). Substitute: \(y = (x + 3)^2 - 9 + 11 = (x + 3)^2 + 2\). The distractor \((x - 3)^2 + 2\) is a sign error; because \((x + 3)^2 = (x - (-3))^2\), the vertex is at \(x = -3\).

Discriminant \(= b^2 - 4ac = (5)^2 - 4(1)(7) = 25 - 28 = -3\). The distractor 53 comes from adding instead of subtracting: \(25 + 28 = 53\). Since the discriminant is negative, this equation has no real roots.

A positive discriminant guarantees two distinct real roots. When the discriminant is also a perfect square, its square root is a rational number, so no radical remains in the answer — the roots are rational. Irrational roots arise when the discriminant is positive but not a perfect square.

Divide the exponent by 4 and use the remainder: 5 = 4(1) + 1, so i^5 = i^1 = i. Equivalently, i^5 = i^4 · i = 1 · i = i. The distractor -i is the value of i^3, not i^5.

Add real parts and imaginary parts separately: (4 + (-1)) + (-3 + 7)i = 3 + 4i. The distractor 3 - 4i incorrectly subtracts the imaginary parts instead of adding them. The distractor 5 + 10i adds absolute values without regard to signs.

(2i)(5i) = 10i^2 = 10(-1) = -10. Because i^2 = -1, multiplying two pure imaginary numbers produces a real number. The distractor 10i forgets to apply i^2 = -1 and treats the product as a single imaginary term.

The formula is \((b/2)^2\) where \(b = 5\): \((5/2)^2 = 25/4\). The distractor \(5/2\) takes half of \(b\) without squaring. The distractor 25 squares \(b\) without dividing by 2 first. Note that completing the square often produces fractions when \(b\) is odd.

The vertex is at (-1, -3) and a = 2 > 0, so the parabola opens upward and the vertex is a minimum. The output value is at least -3, giving range y >= -3. If a were negative, the parabola would open downward and the range would be y <= k.

Set the discriminant to zero: \(b^2 - 4(1)(9) = 0\), so \(b^2 = 36\) and \(b = \pm 6\). The distractor \(b = \pm 3\) comes from \(\sqrt{9} = 3\) without multiplying by 4 first. The distractor \(b = \pm 9\) misidentifies \(4ac\) as just \(c = 9\).

Discriminant = (-6)^2 - 4(3)(3) = 36 - 36 = 0. A discriminant of zero means exactly one repeated real root. Factoring confirms: 3x^2 - 6x + 3 = 3(x - 1)^2 = 0, so x = 1 is the only solution. Confusing this with two roots is a common mistake.

Discriminant \(= (-4)^2 - 4(1)(13) = 16 - 52 = -36\). Then \(x = (4 \pm \sqrt{-36}) / 2 = (4 \pm 6i) / 2 = 2 \pm 3i\). The distractor \(2 \pm 6i\) forgets to divide \(\sqrt{-36} = 6i\) by \(2a = 2\), keeping the full numerator term.

Expand: (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i + (-1) = 2i. The real parts cancel: 1 + (-1) = 0. The distractor 2 + 2i is the most common error — students compute 1 + 2i + 1 by treating i^2 as +1 instead of -1.

Expand using FOIL: (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i + (-1) = 3 - 4i. The distractor 3 + 4i incorrectly writes +4i for the middle term. The distractor 5 - 4i fails to substitute i^2 = -1, leaving 4 + 1 = 5 instead of 4 - 1 = 3.

Set the discriminant to zero: \(k^2 - 4(1)(16) = 0\), so \(k^2 = 64\) and \(k = \pm 8\). The distractor \(k = \pm 4\) comes from \(\sqrt{16} = 4\) while ignoring the factor of 4 in \(4ac\). The distractor \(k = \pm 16\) squares \(c\) instead of computing \(4ac\) correctly.

Start with y = a(x + 2)^2 + 5. Substitute the point (0, 1): 1 = a(2)^2 + 5 = 4a + 5, so 4a = -4 and a = -1. The equation is y = -(x + 2)^2 + 5. The distractor a = -1/4 comes from incorrectly computing -4 / 16 instead of -4 / 4.

Complete the square: \(x^2 - 6x + 9 = -13 + 9 = -4\), so \((x - 3)^2 = -4\). Taking the square root: \(x - 3 = \pm \sqrt{-4} = \pm 2i\), giving \(x = 3 \pm 2i\). The distractor \(3 \pm 4i\) incorrectly uses \(\sqrt{4} = 4\) instead of \(\sqrt{-4} = 2i\).

The product of complex conjugates (a + bi)(a - bi) = a^2 + b^2 = (2)^2 + (3)^2 = 4 + 9 = 13. By Vieta's formulas, this product equals c/a in the quadratic ax^2 + bx + c. The distractor 4 only squares the real part, and 9 only squares the imaginary part.

In vertex form y = a(x - h)^2 + k, the vertex is (h, k). Here h = 3 and k = 4, so the vertex is (3, 4). A common error is negating h to get (-3, 4), but the expression (x - 3) already signals h = 3 directly.

Rewrite as y = (x - (-2))^2 - 5, so h = -2 and the axis of symmetry is x = -2. Choosing x = 2 is a sign error: the vertex form (x - h)^2 means (x + 2) corresponds to h = -2, not h = 2.

By definition, \(i = \sqrt{-1}\), so \(i^2 = -1\). This is the foundational identity for complex numbers. Confusing \(i^2\) with \(i^0 = 1\) or leaving it as \(i\) are common errors when students first encounter the imaginary unit.

A complex number in standard form is a + bi, where a is the real part and b is the imaginary part. In 6 - 9i, the real part is 6 and the imaginary part is -9. The imaginary part is the coefficient of i, not the entire term.

Vertex form is y = a(x - h)^2 + k, where (h, k) is the vertex. Standard form is y = ax^2 + bx + c, and factored form is y = a(x - r)(x - s) where r and s are x-intercepts. Each form highlights different features of the parabola.

The direction a parabola opens is determined solely by the sign of a. Here a = -1 < 0, so the parabola opens downward and has a maximum. The position of the vertex and the sign of the constant have no bearing on the direction of opening.

Since a = 1 > 0, the parabola opens upward and reaches a minimum at the vertex (4, 7). The minimum value of y is k = 7. Note that x = 4 is the input where the minimum occurs, not the minimum output value itself.

\(\sqrt{-36} = \sqrt{36 \cdot (-1)} = \sqrt{36} \cdot \sqrt{-1} = 6 \cdot i = 6i\). By convention, the principal square root takes the positive value, so the result is \(6i\) rather than \(-6i\). The result is imaginary, not real.

Complete the square: take half of \(-6\) to get \(-3\), square it to get 9, then rewrite as \(y = (x^2 - 6x + 9) + 11 - 9 = (x - 3)^2 + 2\). Choice A has the wrong sign inside the parentheses. Choice C subtracts 9 twice instead of compensating by adding 9 back.

Distribute the subtraction sign: (5 + 2i) - (3 - 4i) = 5 + 2i - 3 + 4i = (5 - 3) + (2 + 4)i = 2 + 6i. The critical step is that subtracting -4i yields +4i. Choice A incorrectly subtracts the imaginary part instead of adding it.

Use FOIL: (3)(2) + (3)(-i) + (i)(2) + (i)(-i) = 6 - 3i + 2i - i^2 = 6 - i - (-1) = 7 - i. The key step is replacing i^2 with -1, which increases the real part by 1. Choice A (6 - i) forgets this correction from i^2.

The discriminant is b^2 - 4ac with a = 2, b = -3, c = 5. Compute (-3)^2 - 4(2)(5) = 9 - 40 = -31. Choice C (-11) results from using a = 1 instead of 2. Choice D (1) comes from computing b^2 - 4a instead of b^2 - 4ac.

Since a = -2 < 0, the parabola opens downward and reaches a maximum at the vertex. In vertex form y = a(x - h)^2 + k, the vertex is (-3, 8), so the maximum y-value is k = 8. The value -3 is the x-coordinate of the vertex, not the maximum output.

Powers of i repeat with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Divide: 10 = 4(2) + 2, so i^10 = i^2 = -1. Equivalently, i^10 = (i^4)^2 * i^2 = 1 * (-1) = -1. The remainder when dividing the exponent by 4 determines the result.

Move the constant: \(x^2 + 8x = -7\). Add \((8/2)^2 = 16\) to both sides: \((x + 4)^2 = 9\). Take square roots: \(x + 4 = \pm 3\). So \(x = -1\) or \(x = -7\). Both roots are negative because the sum of the roots equals \(-8\) (a negative number) for this equation.

When b^2 - 4ac < 0, the quadratic formula produces sqrt of a negative number, yielding two complex non-real conjugate solutions of the form p ± qi where q is not 0. Choice C is incorrect: the solutions do have a real part equal to -b/(2a), which is generally nonzero.

Set y = 0: 2(x - 1)^2 = 8, so (x - 1)^2 = 4, giving x - 1 = ±2. Thus x = 3 or x = -1. Choice D (4 and -2) results from forgetting to divide by 2 first, taking the square root of 8 directly instead of 4.

Multiply numerator and denominator by the conjugate (1 + 2i): numerator = (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i; denominator = 1^2 + 2^2 = 5. Result: (-5 + 10i)/5 = -1 + 2i. Multiplying by the conjugate eliminates the imaginary part of the denominator.

For two distinct real roots, the discriminant must be strictly positive: \(b^2 - 4ac > 0\). With \(a = 1\) and \(b = -10\): \(100 - 4c > 0\), so \(c < 25\). At \(c = 25\) the discriminant is zero (one repeated root); for \(c > 25\) the roots are non-real complex numbers.

Sum of roots \(= (3 + i) + (3 - i) = 6\); product of roots \(= (3 + i)(3 - i) = 9 - i^2 = 9 + 1 = 10\). A monic quadratic is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - 6x + 10\). Choice A has the wrong sign on the linear term; choice B has the wrong sign on the constant.

The maximum occurs at the vertex: t = -b/(2a) = -64 / (2 * (-16)) = 2 seconds. Evaluate: h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69 feet. Choice A (64) is the initial upward velocity coefficient, not the maximum height. Choice D incorrectly adds the initial height to the launch velocity.

With \(a = 2\), \(b = 1\), \(c = 3\): discriminant \(= 1 - 4(2)(3) = 1 - 24 = -23\). So \(x = (-1 \pm \sqrt{-23}) / 4 = (-1 \pm i\sqrt{23}) / 4\). Choice A omits the imaginary unit \(i\). Choice D uses denominator 2 instead of \(2a = 4\).

Multiply numerator and denominator by the conjugate (2 + 3i): numerator = (2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i; denominator = 2^2 + 3^2 = 13. Result: -5/13 + (12/13)i. Choice A (1) is incorrect; a complex number divided by its conjugate equals 1 only if the modulus is 1.

Substitute the vertex (h, k) = (2, -3) and the point (4, 5) into vertex form: 5 = a(4 - 2)^2 - 3 = 4a - 3. Solving gives 4a = 8, so a = 2. Choice A (a = 1/2) would produce y = 2(0.25) - 3 = -2.5 at x = 4, not 5.

For a quadratic with real coefficients, complex roots come in conjugate pairs, so the other root is \(1 - 2i\). Sum of roots \(= 2\); product \(= (1 + 2i)(1 - 2i) = 1 + 4 = 5\). The equation is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - 2x + 5 = 0\). Choice B has the wrong sign on the linear term.

In vertex form y = a(x - h)^2 + k, the vertex is (h, k). Here the expression is (x - 4), which means h = +4, and k = 7, giving the vertex (4, 7). Choice A is a common error: students see the minus sign in (x - 4) and write h = -4, but the form subtracts h, so (x - 4) means h = 4.

By definition, \(i\) is the imaginary unit where \(i = \sqrt{-1}\), so \(i^2 = -1\). This is the foundational property of complex numbers. Choice A (1) is incorrect — that is the value of \(i^0\) or \(i^4\), not \(i^2\).

In vertex form y = a(x - h)^2 + k, the axis of symmetry is x = h. Rewriting: y = 3(x - (-2))^2 - 5 shows h = -2. So the axis of symmetry is x = -2. Choice B (x = 2) is wrong because (x + 2) must be read as (x - (-2)), making h = -2, not +2.

The discriminant controls the number and type of roots. When b^2 - 4ac = 0, the quadratic formula reduces to x = -b / (2a), producing a single value called a repeated or double root. Choice A requires discriminant greater than 0; Choice B requires discriminant less than 0.

Every complex number in standard form a + bi has a as the real part and b as the imaginary part. For 5 - 3i, the real part is 5 and the imaginary part is -3. Choice A (-3) is the imaginary part, not the real part.

Vertex form is \(y = a(x - h)^2 + k\). Choice B, \(y = (x + 3)^2\), matches this pattern with \(a = 1\), \(h = -3\), and \(k = 0\). Choice A is written in standard form (even though it happens to be a perfect square trinomial). Choice C is also standard form.

Powers of i follow a cycle of 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. You can also compute directly: i^4 = (i^2)^2 = (-1)^2 = 1. Choice A (-1) is the value of i^2, not i^4.

Complete the square: \(y = x^2 - 6x + 11 = (x^2 - 6x + 9) + 11 - 9 = (x - 3)^2 + 2\). The vertex is \((3, 2)\). Choice C is wrong because \(11 - 9 = +2\), not \(-2\). Choice B uses \(h = -3\), which would require \((x + 3)\), but \((x - 3)\) gives \(h = +3\).

Add real parts and imaginary parts separately: (4 + 2) + (3 + (-5))i = 6 + (-2)i = 6 - 2i. Choice C (6 + 8i) incorrectly adds the absolute values of the imaginary parts (3 + 5 = 8) rather than treating -5i as a subtraction.

With \(a = 1\), \(b = 6\), \(c = 9\): discriminant \(= 6^2 - 4(1)(9) = 36 - 36 = 0\). A discriminant of 0 means exactly one real solution (a repeated root). Indeed, \(x^2 + 6x + 9 = (x + 3)^2 = 0\) gives \(x = -3\) as a double root. Choice D is wrong: discriminant \(= 0\) means one real root, not no real roots.

Rearrange: \(x^2 - 4x = 12\). Half of \(-4\) is \(-2\), and \((-2)^2 = 4\). Add 4 to both sides: \(x^2 - 4x + 4 = 16\), giving \((x - 2)^2 = 16\). Taking square roots: \(x - 2 = \pm 4\), so \(x = 6\) or \(x = -2\). Choice D uses \((x + 2)^2\), which would require the middle term to be \(+4x\), not \(-4x\).

The leading coefficient a determines whether the parabola opens upward (a > 0) or downward (a < 0). The values h and k only shift the vertex location and do not affect the direction of opening. Choice D (a > 0) describes an upward-opening parabola.

Complete the square: \(y = (x^2 + 10x + 25) + 21 - 25 = (x + 5)^2 - 4\). The vertex is \((-5, -4)\). Alternatively, \(x = -\frac{b}{2a} = -\frac{10}{2} = -5\), then \(y = 25 - 50 + 21 = -4\). Choice C has the right x-coordinate but the wrong sign for the k-value.

Divide the exponent by 4 to find the remainder: 23 = 4 times 5 + 3, so the remainder is 3. Therefore i^23 = i^3 = -i. The cycle is i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Choice C (i) would be correct for an exponent with remainder 1.

Since a = 2 > 0, the parabola opens upward and has a minimum at its vertex. The vertex is at (3, 1), so the minimum value of y is 1. Choice A (3) is the x-coordinate of the vertex, not the minimum y-value.

Distribute the minus sign: (6 - 2i) - (3 + 4i) = 6 - 2i - 3 - 4i = (6 - 3) + (-2 - 4)i = 3 - 6i. Choice A (3 + 2i) is wrong because subtracting +4i from -2i gives -6i, not +2i.

With \(a = 3\), \(b = 6\), \(c = 3\): discriminant \(= b^2 - 4ac = 36 - 4(3)(3) = 36 - 36 = 0\). This means one repeated root. Dividing the equation by 3 gives \(x^2 + 2x + 1 = (x + 1)^2 = 0\), confirming \(x = -1\) is a double root. Choice A (72) incorrectly computes \(4ac\) as something other than 36.

To complete the square for \(x^2 + bx\), add \((b/2)^2\). Here \(b = 12\), so \((12/2)^2 = 6^2 = 36\). This gives \(x^2 + 12x + 36 = (x + 6)^2\). Choice A (6) is only half of 12 — you must square that half-value to get the correct constant.

For exactly one real solution, set the discriminant equal to zero. With a = k, b = -4, c = 1: (-4)^2 - 4(k)(1) = 0, so 16 - 4k = 0 and k = 4. Choice C (k = 16) gives discriminant 16 - 64 = -48 < 0, which produces no real solutions, not one.

Start with vertex form using h = -1 and k = 4: y = a(x + 1)^2 + 4. Substitute the point (0, 2): 2 = a(0 + 1)^2 + 4 = a + 4, so a = -2. The equation is y = -2(x + 1)^2 + 4. Choice D (a = +2) gives y(0) = 2(1) + 4 = 6, which does not pass through (0, 2).

Expand using FOIL: (1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2 = 1 + 4i + 4i^2 = 1 + 4i + 4(-1) = 1 - 4 + 4i = -3 + 4i. Choice C (3 + 4i) is a common error from treating i^2 as +1 instead of -1, giving 1 + 4 = 5... wait that would be 5 + 4i, not 3 + 4i. Choice C arises from computing (2i)^2 = +4 and getting 1 + 4 + 4i = 5 + 4i — actually choice C likely comes from forgetting to apply i^2 = -1 to get -4 and instead using just 4i^2 without substituting, resulting in sign errors.

Two distinct real roots require a positive discriminant: \(k^2 - 4(1)(4) > 0\), so \(k^2 - 16 > 0\), meaning \(k^2 > 16\). This gives \(k > 4\) or \(k < -4\). Choice A only captures the positive case, missing \(k < -4\). Choice D (\(k = 4\) or \(k = -4\)) gives discriminant \(= 0\), which is one repeated root, not two distinct roots.

Apply the quadratic formula with \(a = 2\), \(b = -4\), \(c = 5\): \(x = \frac{4 \pm \sqrt{16 - 40}}{4} = \frac{4 \pm \sqrt{-24}}{4} = \frac{4 \pm 2i\sqrt{6}}{4} = 1 \pm \frac{\sqrt{6}}{2}i\). Choice B incorrectly simplifies \(2i\sqrt{6}/4\) as \(\sqrt{6}i\) instead of \(\frac{\sqrt{6}}{2}i\), failing to divide the radical by 2.

Substitute the point (4, 3) into the vertex form: 3 = a(4 - 2)^2 - 5 = a(4) - 5 = 4a - 5. Solving: 4a = 8, so a = 2. Since a = 2 > 0, the parabola indeed opens upward as required. Choice C (a = 1/2) gives y(4) = (1/2)(4) - 5 = 2 - 5 = -3, which does not match the given point.

Using the difference of squares pattern: (5 + 2i)(5 - 2i) = 5^2 - (2i)^2 = 25 - 4i^2 = 25 - 4(-1) = 25 + 4 = 29. Choice A (21) results from the error 25 - 4 = 21, which forgets that i^2 = -1 turns the subtraction into addition. The product of any complex number and its conjugate always yields a real number equal to a^2 + b^2.

In vertex form y = (x - h)^2 + k, the vertex is (h, k). Here (x - 4) means h = 4, and k = 7, so the vertex is (4, 7). A common error is to read the sign directly from inside the parentheses and write h = -4; since the form is (x - h), the sign flips.

Rewrite (x + 6) as (x - (-6)), which gives h = -6. The axis of symmetry is always the vertical line x = h, so x = -6. Choosing x = 6 is a sign error from reading the number inside the parentheses without accounting for the subtraction in the standard vertex form.

By definition, i is the imaginary unit satisfying i^2 = -1. This is the foundational property of complex numbers. The powers of i cycle with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Choosing 1 confuses i^2 with i^4.

\(\sqrt{-25} = \sqrt{25} \cdot \sqrt{-1} = 5 \cdot i = 5i\). The imaginary unit \(i = \sqrt{-1}\) converts a negative radicand into a real result times \(i\). The principal square root is positive, so the answer is \(5i\), not \(-5i\). Choosing \(25i\) incorrectly uses \(-25\) as the coefficient without taking the square root of 25.

The vertex is the point (h, k). When x = h, the squared term equals zero and y = k, making k the y-coordinate of the vertex. The variable a determines the direction and width of the parabola, and h is the x-coordinate of the vertex.

When the discriminant is zero, the quadratic formula gives x = -b / (2a), a single repeated root. Geometrically, the parabola is tangent to the x-axis at exactly one point. A negative discriminant yields zero real solutions, and a positive discriminant yields two distinct real solutions.

A complex number written as a + bi has real part a and imaginary part b. For -5 + 3i, the real part is -5 and the imaginary coefficient is 3. Choosing 3 confuses the imaginary coefficient with the real part.

With \(a = 1\), \(b = -5\), \(c = 6\): discriminant \(= b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1\). Because the discriminant is positive, there are two distinct real roots. Choosing 25 omits the subtraction of \(4ac\). Choosing 49 results from adding \(4ac\) instead of subtracting: \(25 + 24 = 49\).

To complete the square, take half the coefficient of \(x\) and square it: \((8/2)^2 = 4^2 = 16\). Adding 16 gives \(x^2 + 8x + 16 = (x + 4)^2\). A common mistake is to add 4 (half of 8 without squaring) or to add 64 (squaring 8 without first halving it).

Complete the square: \(x^2 - 6x + 5 = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4\). The vertex is \((3, -4)\). Choosing \((x + 3)^2 - 4\) reverses the sign of \(h\). Choosing \((x - 3)^2 + 4\) incorrectly adds the correction term instead of subtracting it.

Add real parts and imaginary parts separately: (3 + 1) + (2 + (-5))i = 4 + (-3)i = 4 - 3i. A common mistake is to add absolute values of the imaginary coefficients without considering the sign: 2 + 5 = 7 incorrectly yields 4 + 7i.

Since a = -3 < 0, the parabola opens downward and reaches its maximum at the vertex. The vertex is (h, k) = (2, 6), so the maximum value of y is 6. Choosing -3 mistakes the leading coefficient for the maximum, and choosing 2 gives the x-coordinate of the vertex rather than the y-coordinate.

Use FOIL: (2)(3) + (2)(-2i) + (i)(3) + (i)(-2i) = 6 - 4i + 3i - 2i^2. Since i^2 = -1, the last term becomes -2(-1) = 2. Combining: (6 + 2) + (-4 + 3)i = 8 - i. Forgetting to replace i^2 with -1 leaves the product as 6 - i, which is choice A.

With a = 2, b = -3, c = 5: discriminant = (-3)^2 - 4(2)(5) = 9 - 40 = -31. A negative discriminant means there are no real solutions; the equation has two complex conjugate roots. Choice B results from using a = 1 instead of a = 2: 9 - 4(1)(5) = -11.

Powers of i cycle with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Divide the exponent: 10 = 4(2) + 2, so i^10 = (i^4)^2 * i^2 = 1 * (-1) = -1. Choosing 1 confuses i^10 with i^8, which does equal 1 since 8 is divisible by 4.

An upward-opening parabola requires a > 0, and vertex (2, 3) requires h = 2 and k = 3. Choice C fits: y = (x - 2)^2 + 3 has a = 1 > 0 and vertex (2, 3). Choice A has a = -1, so it opens downward. Choice B has vertex (-2, 3) due to the sign error in h. Choice D places the vertex at (2, -3).

Find h using h = -b / (2a) = -6 / (2 * 3) = -1. Substitute x = -1 to find k: k = 3(-1)^2 + 6(-1) - 1 = 3 - 6 - 1 = -4. The vertex is (-1, -4). Choice D results from the common mistake of using h = -b / a = -6 / 3 = -2, omitting the factor of 2 in the denominator.

Multiply numerator and denominator by the conjugate of the denominator, (1 - i): numerator becomes (2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 2 + i + 3 = 5 + i; denominator becomes (1)^2 + (1)^2 = 2. Result: (5 + i) / 2 = 5/2 + (1/2)i. Choice D results from multiplying by (1 + i) instead of the conjugate (1 - i).

Exactly one real solution requires the discriminant to equal zero: \(b^2 - 4(1)(25) = 0\), so \(b^2 = 100\), giving \(b = \pm 10\). Both values work: \(x^2 + 10x + 25 = (x + 5)^2\) and \(x^2 - 10x + 25 = (x - 5)^2\). Choosing \(\pm 5\) incorrectly uses \(\sqrt{25}\) rather than solving \(b^2 = 100\). Only listing \(b = 10\) ignores the symmetric negative case.

Build the quadratic from its roots: \((x - (-2 + i))(x - (-2 - i)) = (x + 2 - i)(x + 2 + i) = (x + 2)^2 - i^2 = x^2 + 4x + 4 - (-1) = x^2 + 4x + 5\). Alternatively, the sum of roots is \(-4\) so the linear coefficient is \(+4\), and the product of roots is \((-2)^2 + 1^2 = 5\). Choice A incorrectly uses \(-4x\) for the linear term.

Square in two steps. First: (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i. Then: (2i)^2 = 4i^2 = 4(-1) = -4. Choice A forgets that i^2 = -1, giving 4 instead of -4. Choice D stops at the intermediate result (1 + i)^2 = 2i without squaring again.

Factor 3 from the first two terms: \(3(x^2 - 4x) + 7\). Complete the square: \(3(x^2 - 4x + 4 - 4) + 7 = 3(x - 2)^2 - 12 + 7 = 3(x - 2)^2 - 5\). Choice B omits the \(-12\) correction, incorrectly leaving \(+7\). Choice D adds the correction inside but forgets to add 7 back, stopping at \(3(x - 2)^2 - 12\).

Substitute the point (0, 1): 1 = a(0 + 2)^2 + 5 = 4a + 5. Solving: 4a = -4, so a = -1. The negative value is consistent with the parabola opening downward, since the given point y = 1 is below the vertex y = 5. Choosing a = 1 would give y = 4(1) + 5 = 9 at x = 0, not 1.

The discriminant is \(4^2 - 4(1)(13) = 16 - 52 = -36\). Using the quadratic formula: \(x = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i\). Choice C forgets to divide by 2, giving \(\pm 6i\) instead of \(\pm 3i\). Choice A uses the wrong sign for the real part, writing \(+2\) instead of \(-2\).

The coefficient \(a = -2\) causes two effects: the negative sign reflects the graph over the x-axis, and \(|a| = 2 > 1\) stretches it vertically by factor 2 (a compression would require \(0 < |a| < 1\)). The expression \((x + 3)\) equals \((x - (-3))\), so \(h = -3\) and the shift is 3 units left, not right. Finally \(k = 5\) shifts the graph up 5 units. Choice C incorrectly states the horizontal shift direction.

By definition, i is the imaginary unit where i = √(-1), so i^2 = -1. This is the foundational property of imaginary numbers. The distractor 1 is incorrect — i^4 = 1, not i^2. The cycle of powers is i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1.

In vertex form y = (x - h)^2 + k, the vertex is (h, k). Here h = 5 and k = 3, giving vertex (5, 3). A common error is choosing (-5, 3) by not accounting for the sign in (x - h): since the expression is (x - 5), h = +5, not -5.

Rewrite in standard vertex form: y = 4(x - (-3))^2 - 7, so h = -3. The axis of symmetry is always x = h, giving x = -3. The distractor x = 3 results from ignoring the sign: (x + 3) means h = -3, not +3.

When a < 0, the squared term is negated, so the parabola opens downward. A downward-opening parabola reaches a highest point called a maximum, not a minimum. When a > 0, the parabola opens upward and has a minimum. A downward-opening parabola extends infinitely downward and cannot have a minimum.

√(-36) = √(36 × (-1)) = √36 × √(-1) = 6 × i = 6i. The principal square root is positive, so the answer is 6i, not -6i. Choices 6 and -6 are real numbers, but the square root of a negative number is not a real number — it requires the imaginary unit i.

For a complex number a + bi, the imaginary part is b — the coefficient of i, not including the i itself. For 7 - 2i, b = -2, so the imaginary part is -2 (a real number). The distractor -2i includes the unit i, which makes it the imaginary term, not the imaginary part. The value 7 is the real part.

The discriminant b^2 - 4ac determines the nature of roots: positive means two distinct real roots, zero means one repeated real root, and negative means two complex conjugate roots. The x-coordinate of the vertex is -b/(2a), the y-intercept is the value c (when x = 0), and the sum of the solutions is -b/a by Vieta's formulas.

Subtract real and imaginary parts separately: real part = 4 - 2 = 2; imaginary part = -3 - 5 = -8. Result: 2 - 8i. A common error is adding the imaginary parts instead of subtracting: -3 + 5 = 2, yielding 2 + 2i. Another error is adding the real parts: 4 + 2 = 6, yielding 6 - 8i.

To complete the square, take half the coefficient of \(x\) and square it: \((10/2)^2 = 5^2 = 25\). Then \(x^2 + 10x + 25 = (x + 5)^2\). The distractor 100 comes from squaring 10 directly without halving first. The distractor 5 comes from halving without squaring. The value 20 has no systematic basis.

When b^2 - 4ac = 0, the quadratic formula gives x = -b/(2a), a single value — called a repeated or double root. The parabola is tangent to the x-axis at exactly one point. A negative discriminant gives two complex conjugate solutions; a positive discriminant gives two distinct real solutions.

Complete the square: \(x^2 - 6x + 11 = (x^2 - 6x + 9) - 9 + 11 = (x - 3)^2 + 2\). The vertex is \((3, 2)\). Choice \((x + 3)^2 + 2\) has the wrong sign for \(h\). Choice \((x - 3)^2 - 9\) omits the original constant \(+11\). Choice \((x - 6)^2 - 25\) incorrectly uses 6 as \(h\) without halving the coefficient.

The modulus of a + bi is √(a^2 + b^2) = √(25 + 144) = √169 = 13. This uses the Pythagorean triple (5, 12, 13). The distractor 17 comes from simply adding 5 + 12. The distractor 169 is the value before taking the square root. Always take the square root at the final step.

The complex conjugate of a + bi is a - bi: only the sign of the imaginary part changes; the real part stays the same. For -4 + 9i, the conjugate is -4 - 9i. The distractor 4 - 9i incorrectly negates the real part as well. The conjugate is used to eliminate the imaginary part when dividing complex numbers.

Powers of i cycle with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Divide 19 by 4: 19 = 4 × 4 + 3, giving remainder 3. Therefore i^19 = i^3 = -i. The distractor i corresponds to remainder 1 (that would be i^17). A remainder of 0 gives 1, and a remainder of 2 gives -1.

Since a = -2 < 0, the parabola opens downward and the vertex at (1, 6) is the maximum point. The output can be at most 6 and decreases without bound, so the range is y ≤ 6. The range y ≥ 6 describes upward-opening parabolas. The value -2 is the leading coefficient, not a boundary for the output.

These are complex conjugates, so (a + bi)(a - bi) = a^2 + b^2 = 4 + 9 = 13. Expanding directly: 4 - 6i + 6i - 9i^2 = 4 - 9(-1) = 13. The distractor -5 results from computing 4 - 9 = -5, incorrectly treating i^2 as +1 instead of -1. Multiplying conjugate pairs always yields a real number.

Compute the discriminant: \(b^2 - 4ac = 9 - 4(1)(5) = 9 - 20 = -11\). Since the discriminant is negative, \(x^2 + 3x + 5 = 0\) has no real solutions, so the parabola never crosses the x-axis — it has 0 x-intercepts. With \(a = 1 > 0\), the parabola opens upward and its minimum sits entirely above the x-axis.

In vertex form y = a(x - h)^2 + k with a = 3 > 0, the parabola opens upward and the vertex is the minimum point. The vertex is at h = -1 and k = -7, so the minimum value of y is k = -7. The value 3 is the leading coefficient (vertical stretch), and -1 is the x-coordinate of the vertex — neither is the minimum y-value.

Factor 2 from the quadratic terms: \(y = 2(x^2 - 6x) + 7\). Complete the square inside: add and subtract 9, giving \(y = 2(x^2 - 6x + 9) - 2(9) + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11\). Vertex: \((3, -11)\). The critical error many students make is subtracting 9 instead of \(2(9) = 18\), forgetting to account for the leading coefficient outside the parentheses.

Two distinct real solutions require a strictly positive discriminant: \(b^2 - 4ac > 0\). With \(a = 1\), \(b = -8\), \(c = k\): \(64 - 4k > 0\), so \(k < 16\). If \(k = 16\), the discriminant equals 0 — one repeated root, not two distinct roots. Therefore \(k \le 16\) is too broad for the strict requirement of two distinct solutions.

Multiply numerator and denominator by the conjugate of the denominator, (1 - 2i). Numerator: (3 - i)(1 - 2i) = 3 - 6i - i + 2i^2 = 3 - 7i + 2(-1) = 1 - 7i. Denominator: (1 + 2i)(1 - 2i) = 1 + 4 = 5. Result: (1 - 7i)/5. The distractor (1 + 7i)/5 arises from a sign error when combining -6i - i = -7i — some students mistakenly get +7i.

Write vertex form with h = -1, k = 4: y = a(x + 1)^2 + 4. Substitute the point (1, -4): -4 = a(2)^2 + 4 = 4a + 4, giving 4a = -8 and a = -2. The equation is y = -2(x + 1)^2 + 4. The negative a is consistent — the vertex is the highest point (y = 4) while the parabola passes through a lower point (y = -4). Choice y = -2(x - 1)^2 + 4 uses h = +1 instead of h = -1.

For quadratics with real coefficients, complex roots occur in conjugate pairs. The other solution must be 4 - 3i. The product of the roots is (4 + 3i)(4 - 3i) = 4^2 + 3^2 = 16 + 9 = 25. By Vieta's formulas, this equals c/a. The distractor 8 is the sum of the roots — (4 + 3i) + (4 - 3i) = 8 — not the product. The distractor 16 forgets to add b^2 = 9.

Use the cycle i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. For i^48: 48 ÷ 4 = 12 with remainder 0, so i^48 = i^4 = 1. For i^30: 30 ÷ 4 = 7 with remainder 2, so i^30 = i^2 = -1. Sum: 1 + (-1) = 0. The distractor 2 results from incorrectly computing both powers as 1. A remainder of 0 maps to i^4 = 1, not i^0 — treat remainder 0 as the same as remainder 4.

Write vertex form: y = a(x - 3)^2 + 2. Substitute the origin (0, 0): 0 = a(0 - 3)^2 + 2 = 9a + 2. Solving: 9a = -2, so a = -2/9. The negative sign is consistent — the vertex is at y = 2, which is above the origin at y = 0, so the parabola must open downward. The distractor -9/2 inverts the fraction, as if the equation were solved for 1/a instead of a.

In vertex form y = (x - h)^2 + k, the vertex is (h, k). Here h = 3 and k = 5, giving vertex (3, 5). A common error is reading (x - 3) as h = -3 due to sign confusion — the subtraction inside the parentheses means h is positive 3.

By definition, \(i = \sqrt{-1}\), so \(i^2 = -1\). This is the foundational property of the imaginary unit. Do not confuse with \(i^4 = 1\) or \(i^1 = i\). Every higher power of \(i\) is derived from this single fact.

In vertex form, the vertex is the point (h, k), so k is the y-coordinate of the vertex. The y-intercept is found by substituting x = 0, which equals ah^2 + k — generally different from k unless h = 0. The vertical stretch factor is a, not k.

Vertex form is \(y = a(x - h)^2 + k\). Choice C matches this structure with \(a = 1\), \(h = 2\), and \(k = 1\). Choice A is standard form, choice B is factored (intercept) form, and choice D is also standard form. Only choice C is already expressed as a squared binomial plus a constant.

The powers of i cycle with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Since the exponent 4 is divisible by 4, i^4 = 1. Do not confuse this with i^2 = -1. Any multiple of 4 as an exponent returns a value of 1.

The axis of symmetry passes through the vertex. Rewriting y = (x - (-2))^2 - 3 reveals h = -2, so the axis of symmetry is x = -2. A common sign error is reading (x + 2) as h = +2 rather than h = -2. The line y = -3 is horizontal and represents the minimum value, not the axis of symmetry.

When the discriminant equals 0, the quadratic formula gives x = -b / (2a), a single repeated root. Graphically, the parabola is tangent to the x-axis at its vertex. A positive discriminant gives 2 distinct real solutions, and a negative discriminant gives 0 real solutions (2 complex solutions instead).

Add real parts and imaginary parts separately: (3 + 1) + (2 + (-5))i = 4 + (-3)i = 4 - 3i. A common error is subtracting the real parts (giving 2) or adding the imaginary parts without accounting for the negative sign on -5i, giving 4 + 7i instead.

Take half of 4, which is 2, and square it: \(2^2 = 4\). Rewrite: \(y = (x^2 + 4x + 4) - 4 + 7 = (x + 2)^2 + 3\). The vertex is \((-2, 3)\). Choice C incorrectly keeps the original constant 7 rather than adjusting it after adding and subtracting 4. Choice B misreads the sign of \(h\).

The discriminant is b^2 - 4ac with a = 2, b = -3, c = 5: (-3)^2 - 4(2)(5) = 9 - 40 = -31. Because the discriminant is negative, the equation has two complex (non-real) solutions. A common error is computing 3^2 = 9 instead of (-3)^2 = 9 — both give the same result here — or miscalculating 4(2)(5) as 20 instead of 40.

Since a = -3 is negative, the parabola opens downward and has a maximum at its vertex. The vertex is (2, 8), so the maximum value is y = 8. Choice B is the x-coordinate of the vertex, not the maximum output. Choice C uses the negative of k, which is incorrect.

Expand using FOIL: (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i + (-1) = 2i. A common error is forgetting the middle term 2i, treating it as 1 + i^2 = 1 - 1 = 0. Another error is forgetting that i^2 = -1, leading to 1 + 2i + 1 = 2 + 2i.

Isolate: \(x^2 = -9\), so \(x = \pm \sqrt{-9} = \pm \sqrt{9} \cdot \sqrt{-1} = \pm 3i\). Choice A incorrectly ignores the negative under the radical, treating \(\sqrt{-9}\) as \(\sqrt{9} = 3\). Choice B confuses \(\sqrt{9} = 3\) with 9 itself.

Since a = 2 > 0, the parabola opens upward. Rewriting as y = 2(x - (-3))^2 + (-5) identifies h = -3 and k = -5, giving vertex (-3, -5). A frequent sign error is reading (x + 3) as h = +3 instead of h = -3, because the vertex form uses (x - h).

Using FOIL: (1)(3) + (1)(-i) + (2i)(3) + (2i)(-i) = 3 - i + 6i - 2i^2 = 3 + 5i - 2(-1) = 3 + 5i + 2 = 5 + 5i. A common error is forgetting that -2i^2 = -2(-1) = +2, instead leaving it as -2 and getting 1 + 5i.

To complete the square, take half the coefficient of \(x\) and square it: \((10/2)^2 = 5^2 = 25\). Then \(x^2 + 10x + 25 = (x + 5)^2\). Choice A is only the half-value (5) without squaring. Choice B squares the full coefficient 10 instead of its half, giving 100.

Distribute the subtraction sign: (5 - 3i) - 2 - 4i = (5 - 2) + (-3 - 4)i = 3 - 7i. A critical error is failing to distribute the negative to the imaginary part, treating -(2 + 4i) as -2 + 4i, which gives 3 + i instead of 3 - 7i.

Factor out 2 from the first two terms: \(y = 2(x^2 - 6x) + 7\). Complete the square: add and subtract 9 inside: \(y = 2(x^2 - 6x + 9) - 18 + 7 = 2(x - 3)^2 - 11\). Vertex is \((3, -11)\). Choice B forgets to subtract the 18 that was introduced, leaving \(+7\) unchanged. Choice D confuses the axis value with \(b\) directly.

Exactly one real solution requires discriminant \(= 0\): \(k^2 - 4(1)(16) = 0\), so \(k^2 = 64\), giving \(k = \pm 8\). Both values work: \(x^2 + 8x + 16 = (x + 4)^2 = 0\) and \(x^2 - 8x + 16 = (x - 4)^2 = 0\). Choice B gives \(k^2 = 16\), which is too small to zero out the discriminant when \(c = 16\).

First, (2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i. Then (3 + 4i)(2 + i) = 6 + 3i + 8i + 4i^2 = 6 + 11i - 4 = 2 + 11i. A common shortcut error is computing 2^3 + i^3 = 8 - i, which ignores the binomial cross terms entirely.

Start with y = a(x - 1)^2 - 4 and substitute (3, 8): 8 = a(3 - 1)^2 - 4 = 4a - 4, so 4a = 12 and a = 3. The equation is y = 3(x - 1)^2 - 4. Choice A gives a = 2: checking (3, 8) yields 2(4) - 4 = 4, not 8. Choice D has the wrong sign on h, placing the vertex at (-1, -4).

Using the quadratic formula with \(a = 1\), \(b = -4\), \(c = 13\): \(x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i\). Choice B forgets to divide by \(2a = 2\), leaving the numerator undivided. Choice C incorrectly computes \(\sqrt{36} = 9\) instead of 6.

Complex roots of equations with real coefficients come in conjugate pairs, so the other solution is 3 + 4i. Their product is (3 - 4i)(3 + 4i) = 9 - (4i)^2 = 9 - (-16) = 25. By Vieta's formulas, this product equals c/a. Choice A adds the real and imaginary parts of one root (3 + 4 = 7) rather than multiplying conjugates.

Non-real solutions require a negative discriminant: b^2 - 4ac < 0, so (4)^2 - 4(2)(k) < 0, giving 16 - 8k < 0, then 8k > 16, so k > 2. Choice A reverses the inequality. Choice D (k > -2) includes values like k = 0, which gives discriminant 16 > 0 — two real solutions — so it is incorrect.

A range of [5, infinity) means the parabola opens upward (a > 0) with minimum value k = 5. Choice B has a = 1 > 0 and k = 5, giving exactly the range [5, infinity). Choice A has a = -2 < 0, so it opens downward with range (-infinity, 5]. Choice C opens upward but has minimum k = -5. Choice D opens downward with maximum 5.

In vertex form y = a(x - h)^2 + k, the axis of symmetry is x = h. Here h = 5, so the axis is x = 5. A common error is choosing x = -5 by misreading (x - 5)^2 as a shift toward negative values, but the subtraction of 5 means h = +5, not -5.

Rewrite in standard vertex form: y = -(x - (-2))^2 + (-7), so h = -2 and k = -7, giving vertex (-2, -7). Choosing (2, -7) is a sign error — writing (x + 2)^2 is equivalent to (x - (-2))^2, so h = -2, not +2.

By definition, the imaginary unit i satisfies i^2 = -1. This is the foundational property of complex numbers. Choosing 1 confuses i^2 with i^4, and choosing i or -i would imply that i is its own square, which is not the case.

The discriminant is b^2 - 4ac, which appears under the radical in the quadratic formula. It determines the nature of the roots. The choice b^2 + 4ac uses the wrong sign, and b^2 - 2ac incorrectly halves the coefficient 4.

The complex conjugate of a - bi is a + bi — only the sign of the imaginary part changes. So the conjugate of 5 - 3i is 5 + 3i. The real part remains 5, ruling out -5 + 3i and -5 - 3i. The choice 3 - 5i incorrectly swaps the real and imaginary parts.

When the discriminant b^2 - 4ac = 0, the quadratic formula gives x = -b / (2a), a single value called a repeated or double root. A positive discriminant yields two distinct real solutions, and a negative discriminant yields two complex non-real solutions. A quadratic always has at least one solution in the complex numbers.

The vertex is (h, k), read directly from the equation. The expression (x - h)^2 equals zero when x = h, which locates the vertex. Choosing (-h, k) is the most common error — students see the minus sign in (x - h)^2 and negate h, but vertex form is already written with that subtraction built in.

Add real parts and imaginary parts separately: (3 + 4) + (5 + (-2))i = 7 + 3i. Choosing 7 - 3i results from treating the imaginary part as 5 - 2 = 3 but then applying a negative sign in error. Choosing -1 + 7i comes from subtracting the real parts instead of adding them.

Use the cyclic pattern: i^1 = i, i^2 = -1, i^3 = i^2 * i = (-1)(i) = -i, i^4 = 1. So i^3 = -i. Choosing -1 confuses i^3 with i^2. Choosing i treats i^3 as if it equals i^1. Knowing this four-step cycle is essential for simplifying any power of i.

With a = 2, b = -3, and c = 5, the discriminant is b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31. A frequent error is computing b^2 + 4ac = 9 + 40 = 49, using the wrong sign. The negative discriminant confirms this equation has two complex non-real solutions.

Complete the square: take half of the linear coefficient (\(4/2 = 2\)) and square it to get 4. Rewrite: \(y = (x^2 + 4x + 4) + 7 - 4 = (x + 2)^2 + 3\). Choosing \((x + 2)^2 - 3\) makes an arithmetic error, computing \(7 - 4\) as \(-3\) instead of \(+3\). Choosing \((x + 4)^2 + 7\) uses \(b\) instead of \(b/2\) inside the binomial.

With \(a = 1\), \(b = -7\), \(c = 10\): discriminant \(= (-7)^2 - 4(1)(10) = 49 - 40 = 9\). Since \(9 > 0\), there are two distinct real solutions. Choosing 'exactly one real solution' misapplies the rule — a discriminant of zero (not 9) produces a repeated root. Choosing 49 forgets to subtract the \(4ac\) term entirely.

Reduce exponents mod 4 using the cycle i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. Because 15 = 4(3) + 3, i^15 = i^3 = -i. Because 16 = 4(4), i^16 = i^4 = 1. Their sum is 1 + (-i) = 1 - i. Choosing 1 + i confuses i^15 with i^1 = i, forgetting to reduce the exponent.

In \(y = a(x - h)^2 + k\): \(h = 4\) shifts right 4 units, \(k = -2\) shifts down 2 units, and \(a = 3\) with \(|a| > 1\) produces a vertical stretch by factor 3. Choosing 'shift left' misreads \((x - 4)^2\) as movement toward negative \(x\). Choosing 'vertical compression' confuses \(a > 1\) (stretch) with \(0 < a < 1\) (compression).

The vertex is at (-1, 9) and a = -4 < 0, so the parabola opens downward. The vertex is a maximum, meaning the output never exceeds 9. The range is y <= 9. Choosing y >= 9 applies the logic for an upward-opening parabola (positive a), where the vertex would be a minimum instead.

Use the conjugate product identity: (a + bi)(a - bi) = a^2 + b^2. So (3 + 4i)(3 - 4i) = 9 + 16 = 25, a real number. Choosing -7 comes from incorrectly treating i^2 = 1, giving 9 - 16 = -7. Choosing -7 + 24i is the result of computing (3 + 4i)^2 instead of the conjugate product.

Set \(y = 0\): \((x - 3)^2 = 16\), so \(x - 3 = \pm 4\). This gives \(x = 3 + 4 = 7\) and \(x = 3 - 4 = -1\). Choosing \(x = \pm 7\) ignores the horizontal shift and solves \(x^2 = 16\) directly. Choosing \(x = \pm 3\) confuses the vertex x-coordinate with the x-intercepts by ignoring the \(-16\).

Compute the discriminant: \(b^2 - 4ac = 3^2 - 4(1)(5) = 9 - 20 = -11\). Since the discriminant is negative, the equation has no real solutions — both solutions are complex non-real numbers. The discriminant is precisely the tool that determines the number of real solutions without fully solving.

Substitute the vertex into vertex form: y = a(x + 1)^2 + 3. Use the point (1, -5): -5 = a(1 + 1)^2 + 3 = 4a + 3, so 4a = -8 and a = -2. Choosing a = 2 makes a sign error when solving for a. Choosing a = -1/2 incorrectly divides -8 by 16 (using (x+1)^2 evaluated as 16) instead of by 4.

Isolate the variable terms: \(x^2 + 6x = -25\). Add \((6/2)^2 = 9\) to both sides: \((x + 3)^2 = -16\). Taking square roots gives \(x + 3 = \pm \sqrt{-16} = \pm 4i\), so \(x = -3 \pm 4i\). Choosing \(x = -3 \pm 4\) forgets that the square root of a negative number is imaginary. Choosing \(x = 3 \pm 4i\) reverses the sign of \(h\).

For exactly one real solution, set the discriminant to zero: \(k^2 - 4(1)(k + 3) = 0\), giving \(k^2 - 4k - 12 = 0\). Factor: \((k - 6)(k + 2) = 0\), so \(k = 6\) or \(k = -2\). Choosing \(k = -6\) or \(k = 2\) reverses both roots. The other options result from errors when expanding or factoring the discriminant equation.

Expand using FOIL: z^2 = (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i + (-1) = 3 - 4i. Choosing 4 - 4i forgets to evaluate i^2, leaving it as zero instead of -1. Choosing 3 + 4i makes a sign error on the middle term -4i. Choosing 5 - 4i incorrectly substitutes i^2 = +1 instead of -1.

Since the coefficients are real, complex roots come in conjugate pairs, so \(4 + i\) is also a root. By Vieta's formulas: sum of roots \(= (4 - i) + (4 + i) = 8 = -b\), giving \(b = -8\). Product of roots \(= (4 - i)(4 + i) = 16 - i^2 = 16 + 1 = 17 = c\). Choosing \(c = 15\) incorrectly computes \(4^2 - 1^2 = 15\), as if applying a real difference of squares rather than using \(i^2 = -1\).

Factor out 2: \(y = 2(x^2 + 6x) + 5\). Complete the square inside: half of 6 is 3, and \(3^2 = 9\). Add and subtract 9 inside: \(y = 2(x^2 + 6x + 9 - 9) + 5 = 2(x + 3)^2 - 18 + 5 = 2(x + 3)^2 - 13\). Choosing \(-18\) forgets to add back the original constant \(+5\). Choosing \(+5\) forgets to subtract the 18 that was introduced when completing the square.

Factor out -16: h = -16(t^2 - 3t) + 4. Complete the square: half of -3 is -3/2 and (-3/2)^2 = 9/4. Add and subtract 9/4 inside: h = -16(t - 3/2)^2 + (-16)(-9/4) + 4 = -16(t - 3/2)^2 + 36 + 4 = -16(t - 3/2)^2 + 40. The maximum height is 40 feet, occurring at t = 3/2 seconds. Choosing 36 forgets to add the initial height of 4 feet to the 36 gained from completing the square.

In vertex form y = a(x - h)^2 + k, the axis of symmetry is x = h. Rewriting the equation as y = (x - (-5))^2 - 3 reveals h = -5, so the axis of symmetry is x = -5. A common error is reading the sign directly from the expression and choosing x = 5, but the vertex form subtracts h, so a '+5' inside the parentheses means h = -5.

Powers of i cycle with period 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, then the pattern repeats. Divide the exponent by 4: 22 = 4(5) + 2, so the remainder is 2, meaning i^22 = i^2 = -1. Choosing 1 is a common error that corresponds to i^20 or i^24, where the remainder is 0.

In vertex form y = a(x - h)^2 + k, the vertex is (h, k). Here h = 6 and k = 11, giving vertex (6, 11). The negative leading coefficient means the parabola opens downward, but this does not affect the coordinates of the vertex. Choosing (-6, 11) is a sign error — because the form subtracts h, a '-6' inside would give h = 6, not -6.

Complete the square: take half of \(-6\), which is \(-3\), then square it to get 9. Add and subtract 9 inside the expression: \(y = (x^2 - 6x + 9) - 9 + 2 = (x - 3)^2 - 7\). Choice A is wrong because it keeps the \(+2\) without subtracting the 9 that was added, giving \(+2\) instead of \(-7\). Choice D incorrectly adds \(9 + 2 = 11\) instead of computing \(-9 + 2 = -7\).

Compute the discriminant: D = b^2 - 4ac = (3)^2 - 4(2)(5) = 9 - 40 = -31. Since D < 0, there are no real solutions — the two solutions are complex conjugates. If D = 0, there would be exactly one repeated real solution; if D > 0, there would be two distinct real solutions. The discriminant always determines the number of real solutions, so answer D is incorrect.

Work in two steps. First, (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i. Then (1 + i)^4 = (2i)^2 = 4i^2 = 4(-1) = -4. Choice B (4) is wrong because it treats i^2 as 1 instead of -1. Choice A (2 + 2i) results from incorrectly squaring the real and imaginary components separately without applying the binomial expansion.

Multiply numerator and denominator by the conjugate of the denominator, (1 + i): numerator becomes (2 + i)(1 + i) = 2 + 2i + i + i^2 = 2 + 3i - 1 = 1 + 3i; denominator becomes (1 - i)(1 + i) = 1 - i^2 = 1 + 1 = 2. The result is (1 + 3i)/2 = 1/2 + (3/2)i. Choice A (1 + 3i) is wrong because the numerator was computed correctly but never divided by 2. Choice D has the correct real part but the wrong sign on the imaginary part.

Exactly one real solution occurs when the discriminant equals zero: b^2 - 4ac = 0. With b = 6 and c = 1: 36 - 4a(1) = 0, so 4a = 36 and a = 9. Checking: a = 3 gives D = 36 - 12 = 24 > 0 (two solutions). Choosing a = 6 may result from incorrectly solving 6 = a rather than 4a = 36.

Quadratics with real coefficients always have complex roots in conjugate pairs, so \(3 - 2i\) is also a root. Multiply the linear factors: \((x - (3 + 2i))(x - (3 - 2i)) = ((x - 3) - 2i)((x - 3) + 2i) = (x - 3)^2 - (2i)^2 = (x - 3)^2 + 4 = x^2 - 6x + 9 + 4 = x^2 - 6x + 13\). Choice A uses \(+7\) instead of \(+13\), which would come from \((x-3)^2 - 2\) rather than \((x-3)^2 + 4\). Choice D produces negative \(c\), indicating real roots, not complex ones.

Factor out 3: \(3(x^2 - 4x) = -7\). Complete the square inside: half of \(-4\) is \(-2\), squared is 4. Add 4 inside and balance by adding \(3(4) = 12\) to the right side: \(3(x - 2)^2 = -7 + 12 = 5\). Divide: \((x - 2)^2 = 5/3\). Take the square root: \(x - 2 = \pm \sqrt{5/3} = \pm \frac{\sqrt{15}}{3}\). So \(x = 2 \pm \frac{\sqrt{15}}{3}\). Choice B results from forgetting to add 12 to the right side and using \(-7 + 7 = 0\) or similar arithmetic, producing \(\sqrt{21}/3\). Choice D uses \(x = 4\) from an error treating the \(12x\) term as a shift of 4 without dividing by 2.