GED Math Quiz: Quadratics
13 questions · exam conditions
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QuadraticsQuestion 1 of 13

Solve the quadratic equation x2−9x+20=0.x^{2}-9x+20=0. Which set contains all real solutions?

{4,5}\{4,5\}
{−4,5}\{-4,5\}
{−4,−5}\{-4,-5\}
{4,−5}\{4,-5\}
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GED Math Quiz

GED Math Quiz: Quadratics

Practice Quadratics in GED Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Quadratics, giving you a quick way to practice the rules, question types, and explanations that matter most for GED Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve the quadratic equation x2−9x+20=0.x^{2}-9x+20=0. Which set contains all real solutions?

  1. {4,5}\{4,5\} (correct answer)
  2. {−4,5}\{-4,5\}
  3. {−4,−5}\{-4,-5\}
  4. {4,−5}\{4,-5\}
Explanation: When you see a quadratic equation in standard form like x2−9x+20=0x^2 - 9x + 20 = 0, you need to find the values of xx that make the equation true. The most reliable approach is factoring. To factor x2−9x+20x^2 - 9x + 20, you need two numbers that multiply to give 20 (the constant term) and add to give -9 (the coefficient of the middle term). Think through factor pairs of 20: 1×20, 2×10, 4×5. Since you need a sum of -9, you want -4 and -5, because (-4) × (-5) = 20 and (-4) + (-5) = -9. This gives you: x2−9x+20=(x−4)(x−5)=0x^2 - 9x + 20 = (x - 4)(x - 5) = 0 Using the zero product property, if (x−4)(x−5)=0(x - 4)(x - 5) = 0, then either x−4=0x - 4 = 0 or x−5=0x - 5 = 0. Solving these gives x=4x = 4 or x=5x = 5. The solution set is {4,5}\{4, 5\}, which is answer choice A. Choice B {−4,5}\{-4, 5\} would result from incorrectly factoring as (x+4)(x−5)(x + 4)(x - 5). Choice C {−4,−5}\{-4, -5\} would come from (x+4)(x+5)(x + 4)(x + 5), which expands to x2+9x+20x^2 + 9x + 20—notice the wrong signs. Choice D {4,−5}\{4, -5\} would result from (x−4)(x+5)=x2+x−20(x - 4)(x + 5) = x^2 + x - 20, which has the wrong middle and constant terms. Always check your factoring by expanding back to the original equation. When factoring ax2+bx+cax^2 + bx + c, remember that the signs in your factors directly determine the signs of your solutions.

Question 2

Solve the equation 2x2+7x−4=02x^2 + 7x - 4 = 0 using the quadratic formula.

  1. x=7±814x = \frac{7 \pm \sqrt{81}}{4}
  2. x=−7±654x = \frac{-7 \pm \sqrt{65}}{4}
  3. x=−7±814x = \frac{-7 \pm \sqrt{81}}{4} (correct answer)
  4. x=−7±494x = \frac{-7 \pm \sqrt{49}}{4}
Explanation: When you encounter a quadratic equation like 2x2+7x−4=02x^2 + 7x - 4 = 0, the quadratic formula is your reliable tool for finding solutions. The formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where you identify the coefficients from the standard form ax2+bx+c=0ax^2 + bx + c = 0. From 2x2+7x−4=02x^2 + 7x - 4 = 0, you can identify: a=2a = 2, b=7b = 7, and c=−4c = -4. Now substitute these values into the quadratic formula: x=−7±72−4(2)(−4)2(2)x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-4)}}{2(2)} Calculate the discriminant (the expression under the square root): 72−4(2)(−4)=49−(−32)=49+32=817^2 - 4(2)(-4) = 49 - (-32) = 49 + 32 = 81 This gives you: x=−7±814x = \frac{-7 \pm \sqrt{81}}{4}, which matches answer choice C. Let's examine why the other options are incorrect. Choice A has 7±814\frac{7 \pm \sqrt{81}}{4} - this incorrectly makes the 7 positive instead of negative, forgetting the negative sign in front of bb. Choice B shows −7±654\frac{-7 \pm \sqrt{65}}{4} - this has the correct sign but calculates the discriminant incorrectly, likely from arithmetic errors when computing 49+3249 + 32. Choice D gives −7±494\frac{-7 \pm \sqrt{49}}{4} - this misses the 4ac4ac term entirely in the discriminant, using only b2=49b^2 = 49. Remember: the quadratic formula requires careful attention to signs, especially the negative in front of bb. Double-check your discriminant calculation since arithmetic errors here derail the entire solution.

Question 3

Solve 5p2=20p.5p^{2}=20p. What are all real solutions for pp?

  1. p=0 or p=4p=0 \text{ or } p=4 (correct answer)
  2. p=14 or p=5p=\dfrac{1}{4} \text{ or } p=5
  3. p=0 or p=14p=0 \text{ or } p=\dfrac{1}{4}
  4. p=4 onlyp=4 \text{ only}
Explanation: When you encounter a quadratic equation like this one, your goal is to find all values of the variable that make the equation true. The key insight here is recognizing that this equation can be solved by factoring. Start by moving all terms to one side: 5p2=20p5p^2 = 20p becomes 5p2−20p=05p^2 - 20p = 0. Now you can factor out the common factor of 5p5p: 5p(p−4)=05p(p - 4) = 0. Using the zero product property, if the product of two factors equals zero, then at least one factor must equal zero. So either 5p=05p = 0 or p−4=0p - 4 = 0. Solving these gives you p=0p = 0 or p=4p = 4. You can verify: when p=0p = 0, both sides equal 0; when p=4p = 4, both sides equal 80. Looking at the wrong answers: Choice B (p=14p = \frac{1}{4} or p=5p = 5) likely comes from algebraic errors during factoring or solving. Choice C (p=0p = 0 or p=14p = \frac{1}{4}) gets the zero correct but makes an error finding the second solution—possibly from incorrectly solving p−4=0p - 4 = 0. Choice D (p=4p = 4 only) finds one correct solution but misses that quadratic equations typically have two solutions, failing to recognize that p=0p = 0 also works. The correct answer is A. Remember: quadratic equations usually have two solutions. When you see a quadratic where one side equals zero, always look for factoring opportunities first—it's often faster than using the quadratic formula.

Question 4

A projectile is launched upward from ground level. Its height h (in feet) after t seconds is given by h=−16t2+64th = -16t^2 + 64t. At what time(s) will the projectile be at a height of 48 feet?

  1. At t = 1 second and t = 3 seconds (correct answer)
  2. At t = 1.5 seconds and t = 2.5 seconds
  3. At t = 2 seconds only
  4. At t = 0.75 seconds and t = 3.25 seconds
Explanation: Set h = 48: -16t² + 64t = 48. Rearranging: -16t² + 64t - 48 = 0. Divide by -16: t² - 4t + 3 = 0. Factoring: (t - 1)(t - 3) = 0, so t = 1 or t = 3 seconds. Choice B results from incorrectly setting up the equation as -16t² + 64t = 60. Choice C comes from finding only the vertex time (maximum height). Choice D results from computational errors in applying the quadratic formula.

Question 5

If 3x2−12x+k=03x^2 - 12x + k = 0 has exactly one solution, what is the value of k?

  1. k = 9
  2. k = 12 (correct answer)
  3. k = 16
  4. k = 24
Explanation: For a quadratic equation to have exactly one solution, the discriminant must equal zero. For 3x² - 12x + k = 0, the discriminant is (-12)² - 4(3)(k) = 144 - 12k. Setting this equal to zero: 144 - 12k = 0, so k = 12. Choice A comes from using the wrong formula b²/4a. Choice C results from forgetting to include the coefficient 3 in the discriminant calculation. Choice D comes from setting the discriminant equal to the coefficient of x.

Question 6

What are the real solutions of 2x2+7x−15=02x^{2}+7x-15=0?

  1. x=−152,  x=1x=-\dfrac{15}{2},\;x=1
  2. x=−5,  x=32x=-5,\;x=\dfrac{3}{2} (correct answer)
  3. x=5,  x=−32x=5,\;x=-\dfrac{3}{2}
  4. x=152,  x=−1x=\dfrac{15}{2},\;x=-1
Explanation: When you encounter a quadratic equation like 2x2+7x−15=02x^{2}+7x-15=0, you're looking for the values of x that make the equation true. You can solve this using factoring, the quadratic formula, or by checking which answer choices work. Let's use the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a} where a=2a=2, b=7b=7, and c=−15c=-15. First, calculate the discriminant: b2−4ac=72−4(2)(−15)=49+120=169b^2-4ac = 7^2-4(2)(-15) = 49+120 = 169 Since 169=13\sqrt{169} = 13, we get: x=−7±134x = \frac{-7 \pm 13}{4} This gives us two solutions: x=−7+134=64=32x = \frac{-7+13}{4} = \frac{6}{4} = \frac{3}{2} and x=−7−134=−204=−5x = \frac{-7-13}{4} = \frac{-20}{4} = -5 So the solutions are x=−5x = -5 and x=32x = \frac{3}{2}, which matches answer choice B. Let's verify why the other choices are wrong. Choice A gives x=−152x = -\frac{15}{2} and x=1x = 1. If you substitute x=1x = 1: 2(1)2+7(1)−15=2+7−15=−6≠02(1)^2 + 7(1) - 15 = 2 + 7 - 15 = -6 \neq 0. Choice C has positive 5 instead of negative 5. Choice D gives x=152x = \frac{15}{2} and x=−1x = -1, but substituting x=−1x = -1: 2(−1)2+7(−1)−15=2−7−15=−20≠02(-1)^2 + 7(-1) - 15 = 2 - 7 - 15 = -20 \neq 0. Always verify your solutions by substituting them back into the original equation. This catches calculation errors and confirms your answer.

Question 7

Using the quadratic formula, solve 3y2+10y+3=03y^{2}+10y+3=0 for all real solutions.

  1. y=−10±646y=\dfrac{-10\pm\sqrt{64}}{6} leading to y=−13 or y=−3y=-\dfrac{1}{3}\text{ or }y=-3 (correct answer)
  2. y=−10±643y=\dfrac{-10\pm\sqrt{64}}{3} giving y=−133 or y=1y=-\dfrac{13}{3}\text{ or }y=1
  3. y=−10±826y=\dfrac{-10\pm\sqrt{82}}{6} no simplification possible
  4. y=−10±1006y=\dfrac{-10\pm\sqrt{100}}{6} giving y=−53 or y=0y=-\dfrac{5}{3}\text{ or }y=0
Explanation: When you encounter a quadratic equation that doesn't factor easily, the quadratic formula is your reliable tool. For any equation in the form ax2+bx+c=0ax^2 + bx + c = 0, the formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. For 3y2+10y+3=03y^2 + 10y + 3 = 0, you have a=3a = 3, b=10b = 10, and c=3c = 3. Substituting these values: y=−10±102−4(3)(3)2(3)=−10±100−366=−10±646y = \frac{-10 \pm \sqrt{10^2 - 4(3)(3)}}{2(3)} = \frac{-10 \pm \sqrt{100 - 36}}{6} = \frac{-10 \pm \sqrt{64}}{6} Since 64=8\sqrt{64} = 8, this gives you y=−10+86=−26=−13y = \frac{-10 + 8}{6} = \frac{-2}{6} = -\frac{1}{3} and y=−10−86=−186=−3y = \frac{-10 - 8}{6} = \frac{-18}{6} = -3. This confirms answer choice A is correct. Choice B uses the wrong denominator (3 instead of 6), forgetting that the denominator should be 2a=2(3)=62a = 2(3) = 6. Choice C incorrectly calculates the discriminant as 82 instead of 64, likely from arithmetic errors in b2−4acb^2 - 4ac. Choice D mistakenly uses 100 as the discriminant, forgetting to subtract 4ac=364ac = 36 from b2=100b^2 = 100. Remember the quadratic formula pattern: identify aa, bb, and cc carefully, then substitute methodically. The denominator is always 2a2a, and double-check your discriminant calculation b2−4acb^2 - 4ac since arithmetic errors here will throw off your entire solution.

Question 8

Solve t2+8t+16=0.t^{2}+8t+16=0. Choose the complete set of real solutions.

  1. t=−4 (double root)t=-4 \text{ (double root)} (correct answer)
  2. t=4 (double root)t=4 \text{ (double root)}
  3. t=−8,  t=2t=-8,\;t=2
  4. t=−2,  t=−8t=-2,\;t=-8
Explanation: When you encounter a quadratic equation like t2+8t+16=0t^2 + 8t + 16 = 0, you have several solving methods available: factoring, using the quadratic formula, or completing the square. The key insight here is recognizing this as a perfect square trinomial. Notice that t2+8t+16t^2 + 8t + 16 follows the pattern a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2. Here, t2t^2 is your a2a^2 term, 16=4216 = 4^2 is your b2b^2 term, and 8t=2⋅t⋅48t = 2 \cdot t \cdot 4 is your 2ab2ab term. This means the expression factors as (t+4)2=0(t + 4)^2 = 0. To solve (t+4)2=0(t + 4)^2 = 0, you take the square root of both sides: t+4=0t + 4 = 0, which gives t=−4t = -4. Since both factors are identical, this is called a double root—the solution t=−4t = -4 has multiplicity 2. Choice B suggests t=4t = 4, but this comes from incorrectly solving t+4=0t + 4 = 0 or factoring with the wrong signs. Choices C and D both show two distinct solutions, which would only occur if the discriminant (b2−4acb^2 - 4ac) were positive. Here, the discriminant is 64−64=064 - 64 = 0, confirming exactly one repeated solution. These incorrect answer choices likely come from factoring errors or misapplying the quadratic formula. Study tip: Always check if a quadratic is a perfect square trinomial before using more complex methods. Perfect squares always produce double roots and follow the pattern a2±2ab+b2=(a±b)2a^2 \pm 2ab + b^2 = (a \pm b)^2.

Question 9

If x2−8x+15=0x^2 - 8x + 15 = 0, what are the solutions?

  1. x = 2 and x = 6
  2. x = -3 and x = -5
  3. x = 1 and x = 15
  4. x = 3 and x = 5 (correct answer)
Explanation: When you encounter a quadratic equation like x2−8x+15=0x^2 - 8x + 15 = 0, you're looking for values of x that make the equation true. The most reliable approach is factoring, where you rewrite the quadratic as a product of two binomials. To factor x2−8x+15x^2 - 8x + 15, you need two numbers that multiply to give 15 (the constant term) and add to give -8 (the coefficient of x). Think through factor pairs of 15: 1×15 and 3×5. Since you need a sum of -8, you want -3 and -5, because (-3) + (-5) = -8 and (-3) × (-5) = 15. This gives you (x−3)(x−5)=0(x - 3)(x - 5) = 0. Using the zero product property, if the product equals zero, then at least one factor must equal zero. So either x−3=0x - 3 = 0 (making x = 3) or x−5=0x - 5 = 0 (making x = 5). This confirms answer D is correct. Let's check why the other options fail. Choice A (x = 2 and x = 6): If you substitute x = 2, you get 4−16+15=3≠04 - 16 + 15 = 3 ≠ 0. Choice B (x = -3 and x = -5): These would come from the equation x2+8x+15=0x^2 + 8x + 15 = 0, not our equation with -8x. Choice C (x = 1 and x = 15): Substituting x = 1 gives 1−8+15=8≠01 - 8 + 15 = 8 ≠ 0. Remember: always verify your solutions by substituting back into the original equation. Both x = 3 and x = 5 should make the equation equal zero.

Question 10

The equation x2+6x=16x^2 + 6x = 16 can be solved by completing the square. What is the correct form after completing the square?

  1. (x+3)2=16(x + 3)^2 = 16
  2. (x+3)2=25(x + 3)^2 = 25 (correct answer)
  3. (x+6)2=52(x + 6)^2 = 52
  4. (x+3)2=7(x + 3)^2 = 7
Explanation: Completing the square is a systematic method for rewriting quadratic equations in perfect square form. When you see an equation like x2+6x=16x^2 + 6x = 16, you're transforming it into (x+h)2=k(x + h)^2 = k format, which makes solving much easier. To complete the square, start with x2+6x=16x^2 + 6x = 16. Take the coefficient of the xx term (which is 6), divide it by 2 to get 3, then square that result: 32=93^2 = 9. Add this value to both sides of the equation: x2+6x+9=16+9x^2 + 6x + 9 = 16 + 9. The left side now factors as a perfect square: (x+3)2=25(x + 3)^2 = 25. Looking at the wrong answers: Choice A gives (x+3)2=16(x + 3)^2 = 16, which means you forgot to add 9 to the right side after adding it to the left. Choice C shows (x+6)2=52(x + 6)^2 = 52, indicating you mistakenly used the entire coefficient (6) instead of half of it (3) inside the parentheses. Choice D gives (x+3)2=7(x + 3)^2 = 7, which would result from subtracting 9 from the right side instead of adding it. The correct answer is B: (x+3)2=25(x + 3)^2 = 25. Remember this pattern: when completing the square for x2+bxx^2 + bx, always take half of bb, square it, and add that result to both sides. The number inside the parentheses is always b2\frac{b}{2}, not the full coefficient.

Question 11

A rectangular garden has an area of 120 square meters. If the length is 2 meters more than twice the width, what is the width of the garden?

  1. 6 meters (correct answer)
  2. 8 meters
  3. 10 meters
  4. 12 meters
Explanation: Let w = width. Then length = 2w + 2. Area = w(2w + 2) = 120. This gives us 2w² + 2w = 120, or w² + w - 60 = 0. Factoring: (w + 8)(w - 6) = 0. Since width must be positive, w = 6 meters. Choice B results from solving 2w² + 2w - 120 = 0 incorrectly. Choice C comes from setting up the equation as w(2w - 2) = 120. Choice D results from confusing width and length in the final answer.

Question 12

The revenue R from selling x units of a product is R=x(50−2x)R = x(50 - 2x). How many units should be sold to maximize revenue?

  1. 15 units
  2. 25 units
  3. 20 units
  4. 12.5 units (correct answer)
Explanation: When you see a revenue function like this, you're dealing with a quadratic optimization problem. Revenue functions often form parabolas that open downward, meaning there's a maximum point at the vertex. To find the maximum revenue, you need to find where the derivative equals zero. First, expand the revenue function: R=x(50−2x)=50x−2x2R = x(50 - 2x) = 50x - 2x^2. Taking the derivative: R′=50−4xR' = 50 - 4x. Setting this equal to zero: 50−4x=050 - 4x = 0, which gives you 4x=504x = 50, so x=12.5x = 12.5 units. You can verify this is a maximum because the second derivative R′′=−4R'' = -4 is negative, confirming the parabola opens downward. Looking at the wrong answers: Choice A (15 units) likely comes from missolving the equation or confusing the setup. Choice B (25 units) might result from incorrectly thinking the maximum occurs at x=50/2=25x = 50/2 = 25 without accounting for the coefficient of the x2x^2 term. Choice C (20 units) could come from arithmetic errors or confusion about the vertex formula. The correct answer is D) 12.5 units because this is where the derivative equals zero, giving the maximum revenue. Study tip: For quadratic optimization problems on the GED, remember that maximums and minimums occur where the derivative equals zero. If you're uncomfortable with calculus, you can also use the vertex formula: for ax2+bx+cax^2 + bx + c, the x-coordinate of the vertex is x=−b/(2a)x = -b/(2a).

Question 13

A quadratic function has roots at x = -2 and x = 4, and passes through the point (0, -8). What is the equation of this function?

  1. y=−x2−2x+8y = -x^2 - 2x + 8
  2. y=x2+2x−8y = x^2 + 2x - 8
  3. y=x2−2x−8y = x^2 - 2x - 8 (correct answer)
  4. y=2x2−4x−16y = 2x^2 - 4x - 16
Explanation: When you encounter a quadratic function problem with given roots and a point, you're working with the factored form and need to find the leading coefficient. Since the quadratic has roots at x = -2 and x = 4, you can write it in factored form as y=a(x+2)(x−4)y = a(x + 2)(x - 4), where 'a' is the leading coefficient you need to determine. The roots tell you where the parabola crosses the x-axis, but you need the additional point (0, -8) to find the exact equation. Substitute the point (0, -8) into the factored form: −8=a(0+2)(0−4)=a(2)(−4)=−8a-8 = a(0 + 2)(0 - 4) = a(2)(-4) = -8a. Solving for 'a': a=1a = 1. So the equation is y=1(x+2)(x−4)y = 1(x + 2)(x - 4). Expanding this: y=x2−4x+2x−8=x2−2x−8y = x^2 - 4x + 2x - 8 = x^2 - 2x - 8, which matches answer C. Let's check why the other options fail. Answer A gives y=−x2−2x+8y = -x^2 - 2x + 8. When x = -2: y=−4+4+8=8≠0y = -4 + 4 + 8 = 8 ≠ 0, so x = -2 isn't actually a root. Answer B gives y=x2+2x−8y = x^2 + 2x - 8. At (0, -8): this works, but when x = -2: y=4−4−8=−8≠0y = 4 - 4 - 8 = -8 ≠ 0, so the roots are wrong. Answer D gives y=2x2−4x−16y = 2x^2 - 4x - 16. At (0, -8): y=−16≠−8y = -16 ≠ -8, so it doesn't pass through the given point. Remember: always use the factored form y=a(x−r1)(x−r2)y = a(x - r_1)(x - r_2) when given roots, then substitute the additional point to find 'a'. This systematic approach prevents calculation errors.