Praxis Math Quiz: Solve Quadratic Equations
20 questions · exam conditions
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Solve Quadratic EquationsQuestion 1 of 20

The solutions to x2−6x+k=0x^2-6x+k=0 are real and equal. What is the value of the solution?

-6
-3
3
9
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Praxis Math Quiz

Praxis Math Quiz: Solve Quadratic Equations

Practice Solve Quadratic Equations in Praxis 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 Solve Quadratic Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Praxis 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

The solutions to x2−6x+k=0x^2-6x+k=0 are real and equal. What is the value of the solution?

  1. -6
  2. -3
  3. 3 (correct answer)
  4. 9
Explanation: If the solutions are real and equal, the quadratic is a perfect square trinomial and the discriminant is 0. The discriminant is b2−4ac=(−6)2−4(1)(k)=36−4kb^2-4ac = (-6)^2-4(1)(k) = 36-4k. Setting this to 0 gives 36−4k=036-4k=0, so k=9k=9. The equation is x2−6x+9=0x^2-6x+9=0, which factors to (x−3)2=0(x-3)^2=0. The only solution is x=3x=3.

Question 2

One solution to the equation x2−11x+k=0x^2 - 11x + k = 0 is x=4x = 4. What is the value of the other solution?

  1. 7 (correct answer)
  2. 11
  3. 15
  4. 28
Explanation: The sum of the two solutions of a quadratic equation ax2+bx+c=0ax^2+bx+c=0 is −b/a-b/a. In this equation, a=1a=1 and b=−11b=-11, so the sum of the solutions is −(−11)/1=11-(-11)/1 = 11. If one solution is 4, let the other solution be r2r_2. Then 4+r2=114 + r_2 = 11. Subtracting 4 from both sides gives r2=7r_2 = 7.

Question 3

If x>0x > 0, what is the larger solution to the equation x+14x=9x + \frac{14}{x} = 9?

  1. 2
  2. 5
  3. 7 (correct answer)
  4. 9
Explanation: To solve the equation, first eliminate the fraction by multiplying every term by xx. This gives x(x)+x(14x)=x(9)x(x) + x(\frac{14}{x}) = x(9), which simplifies to x2+14=9xx^2 + 14 = 9x. Rearrange this into a standard quadratic form: x2−9x+14=0x^2 - 9x + 14 = 0. Factor the quadratic: (x−2)(x−7)=0(x-2)(x-7) = 0. The two solutions are x=2x=2 and x=7x=7. Both are positive. The question asks for the larger solution, which is 7.

Question 4

The company breaks even when its profit is zero. A company's weekly profit PP from selling xx units is given by P(x)=−x2+50x−225P(x) = -x^2 + 50x - 225. The company breaks even at two different production levels. What is the smaller of these two levels?

  1. 5 (correct answer)
  2. 25
  3. 45
  4. 50
Explanation: To find when the company breaks even, set the profit P(x)P(x) to zero: −x2+50x−225=0-x^2 + 50x - 225 = 0. It's easier to solve if we multiply the entire equation by -1, giving x2−50x+225=0x^2 - 50x + 225 = 0. To factor this, we need two numbers that multiply to 225 and add to -50. These numbers are -5 and -45. Thus, the equation factors to (x−5)(x−45)=0(x-5)(x-45) = 0. The two production levels where the company breaks even are x=5x=5 and x=45x=45. The smaller of these two levels is 5.

Question 5

The length of a rectangular garden is 5 feet more than its width. If the area of the garden is 84 square feet, what is the length of the garden in feet?

  1. 7
  2. 12 (correct answer)
  3. 14
  4. 17
Explanation: Let ww represent the width of the garden. The length is w+5w+5. The area of a rectangle is length times width, so w(w+5)=84w(w+5) = 84. Distributing ww gives w2+5w=84w^2 + 5w = 84. To solve this quadratic equation, set it to zero: w2+5w−84=0w^2 + 5w - 84 = 0. Factoring the quadratic, we look for two numbers that multiply to -84 and add to 5. These numbers are 12 and -7. So, the equation factors to (w+12)(w−7)=0(w+12)(w-7) = 0. The possible values for ww are -12 and 7. Since the width cannot be negative, w=7w=7 feet. The question asks for the length, which is w+5=7+5=12w+5 = 7+5 = 12 feet.

Question 6

If one of the solutions to 3x2+bx−8=03x^2+bx-8=0 is 2, what is the other solution?

  1. -4/3 (correct answer)
  2. -2
  3. 4/3
  4. 4
Explanation: Let the two solutions be r1r_1 and r2r_2. The product of the solutions is given by c/ac/a. In this equation, a=3a=3 and c=−8c=-8, so the product is r1⋅r2=−8/3r_1 \cdot r_2 = -8/3. We are given that one solution is r1=2r_1 = 2. So, 2⋅r2=−8/32 \cdot r_2 = -8/3. Dividing by 2, we get r2=−8/6=−4/3r_2 = -8/6 = -4/3.

Question 7

What positive value of xx satisfies the equation 2x2+3x−5=x2−2x+192x^2 + 3x - 5 = x^2 - 2x + 19?

  1. 3 (correct answer)
  2. 4
  3. 6
  4. 8
Explanation: First, bring all terms to one side to form a standard quadratic equation. Subtract x2x^2, add 2x2x, and subtract 1919 from both sides: (2x2−x2)+(3x+2x)+(−5−19)=0(2x^2 - x^2) + (3x + 2x) + (-5 - 19) = 0. This simplifies to x2+5x−24=0x^2 + 5x - 24 = 0. To factor this quadratic, we need two numbers that multiply to -24 and add to 5. These numbers are 8 and -3. So the equation factors to (x+8)(x−3)=0(x+8)(x-3) = 0. The solutions are x=−8x=-8 and x=3x=3. The question asks for the positive value, which is 3.

Question 8

If x2−10x+25=16x^2 - 10x + 25 = 16, which of the following is a possible value for xx?

  1. 1 (correct answer)
  2. 4
  3. 5
  4. 10
Explanation: The expression on the left side, x2−10x+25x^2 - 10x + 25, is a perfect square trinomial, which can be factored as (x−5)2(x-5)^2. So the equation becomes (x−5)2=16(x-5)^2 = 16. Take the square root of both sides: x−5=±4x-5 = \pm 4. This leads to two possibilities: x−5=4x-5 = 4, which gives x=9x=9, or x−5=−4x-5 = -4, which gives x=1x=1. Of the choices provided, 1 is a possible value for xx.

Question 9

For which of the following values of c does the equation 2x2−8x+c=02x^2 - 8x + c = 0 have two distinct real solutions?

  1. 4 (correct answer)
  2. 8
  3. 10
  4. 16
Explanation: An equation has two distinct real solutions if its discriminant, b2−4acb^2 - 4ac, is greater than 0. For the equation 2x2−8x+c=02x^2 - 8x + c = 0, we have a=2a=2, b=−8b=-8, and cc is the unknown. The discriminant is (−8)2−4(2)(c)=64−8c(-8)^2 - 4(2)(c) = 64 - 8c. We need 64−8c>064 - 8c > 0. Adding 8c8c to both sides gives 64>8c64 > 8c. Dividing by 8 gives 8>c8 > c, or c<8c < 8. Of the choices provided, only 4 is less than 8.

Question 10

For the equation y=(x−3)(x+5)y = (x-3)(x+5), for what negative value of xx does y=0y=0?

  1. -15
  2. -5 (correct answer)
  3. -3
  4. -2
Explanation: To find the values of xx for which y=0y=0, we set the equation to zero: (x−3)(x+5)=0(x-3)(x+5) = 0. By the zero product property, either x−3=0x-3=0 or x+5=0x+5=0. Solving these linear equations gives x=3x=3 and x=−5x=-5. The question asks for the negative value of xx, which is -5.

Question 11

The lengths of the two legs of a right triangle are xx and x−7x-7 centimeters. If the hypotenuse is 13 centimeters long, what is the value of xx?

  1. 5
  2. 10
  3. 12 (correct answer)
  4. 15
Explanation: According to the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2. For this triangle, x2+(x−7)2=132x^2 + (x-7)^2 = 13^2. Expanding (x−7)2(x-7)^2 gives x2−14x+49x^2 - 14x + 49. The equation becomes x2+x2−14x+49=169x^2 + x^2 - 14x + 49 = 169. Combining like terms and setting the equation to zero gives 2x2−14x−120=02x^2 - 14x - 120 = 0. Divide the entire equation by 2 to simplify: x2−7x−60=0x^2 - 7x - 60 = 0. Factoring this quadratic, we need two numbers that multiply to -60 and add to -7. These are -12 and 5. So, (x−12)(x+5)=0(x-12)(x+5) = 0. The solutions are x=12x=12 and x=−5x=-5. Since length must be positive, x=12x=12.

Question 12

Which of the following is a solution to the equation (2x+1)2=49(2x+1)^2 = 49?

  1. -4 (correct answer)
  2. -3
  3. 4
  4. 24
Explanation: To solve the equation (2x+1)2=49(2x+1)^2 = 49, take the square root of both sides: 2x+1=±492x+1 = \pm\sqrt{49}, which means 2x+1=±72x+1 = \pm 7. This gives two separate linear equations. Case 1: 2x+1=72x+1 = 7. Subtracting 1 gives 2x=62x = 6, so x=3x=3. Case 2: 2x+1=−72x+1 = -7. Subtracting 1 gives 2x=−82x = -8, so x=−4x=-4. The two solutions are 3 and -4. Of the choices given, -4 is a solution.

Question 13

A quadratic equation with integer coefficients has roots at x=−3x = -3 and x=5/2x = 5/2. Which of the following could be the equation?

  1. x2−2x−15=0x^2 - 2x - 15 = 0
  2. 2x2+x−15=02x^2 + x - 15 = 0 (correct answer)
  3. 2x2−x−15=02x^2 - x - 15 = 0
  4. x2+x−15=0x^2 + x - 15 = 0
Explanation: If the roots are −3-3 and 5/25/2, the factors of the quadratic are (x−(−3))(x - (-3)) and (x−5/2)(x - 5/2), which are (x+3)(x+3) and (x−5/2)(x - 5/2). The equation is (x+3)(x−5/2)=0(x+3)(x - 5/2) = 0. Expanding this gives x2−52x+3x−152=0x^2 - \frac{5}{2}x + 3x - \frac{15}{2} = 0. Combine the x-terms: x2+12x−152=0x^2 + \frac{1}{2}x - \frac{15}{2} = 0. To get integer coefficients as specified, multiply the entire equation by 2: 2x2+x−15=02x^2 + x - 15 = 0.

Question 14

What is one of the solutions to the equation 12x2−14x−34=0\frac{1}{2}x^2 - \frac{1}{4}x - \frac{3}{4} = 0?

  1. -1 (correct answer)
  2. -1/2
  3. 1/2
  4. 3
Explanation: To eliminate the fractions, multiply the entire equation by the least common denominator, which is 4. This yields 4(12x2)−4(14x)−4(34)=4(0)4(\frac{1}{2}x^2) - 4(\frac{1}{4}x) - 4(\frac{3}{4}) = 4(0), which simplifies to 2x2−x−3=02x^2 - x - 3 = 0. This can be factored as (2x−3)(x+1)=0(2x-3)(x+1) = 0. Setting each factor to zero gives 2x−3=0⇒x=3/22x-3=0 \Rightarrow x=3/2 and x+1=0⇒x=−1x+1=0 \Rightarrow x=-1. One of the solutions is -1.

Question 15

The product of two consecutive positive odd integers is 143. What is the larger of the two integers?

  1. 9
  2. 11
  3. 13 (correct answer)
  4. 15
Explanation: Let the two consecutive positive odd integers be nn and n+2n+2. Their product is n(n+2)=143n(n+2) = 143. Expanding this gives n2+2n=143n^2 + 2n = 143. Set the equation to zero: n2+2n−143=0n^2 + 2n - 143 = 0. To factor this, we need two numbers that multiply to -143 and add to 2. These numbers are 13 and -11. So, (n+13)(n−11)=0(n+13)(n-11) = 0. The possible values for nn are -13 and 11. Since the integers must be positive, n=11n=11. The two integers are 11 and 11+2=1311+2=13. The larger of the two is 13.

Question 16

What is the largest real solution to the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0?

  1. 2
  2. 3 (correct answer)
  3. 4
  4. 9
Explanation: This equation is quadratic in form. Let u=x2u = x^2. The equation becomes u2−13u+36=0u^2 - 13u + 36 = 0. Factoring this quadratic gives (u−4)(u−9)=0(u-4)(u-9) = 0. So, the solutions for uu are u=4u=4 and u=9u=9. Since u=x2u = x^2, we have two cases: x2=4x^2 = 4, which gives x=±2x = \pm 2, and x2=9x^2 = 9, which gives x=±3x = \pm 3. The set of all real solutions is {−3,−2,2,3}\{-3, -2, 2, 3\}. The largest of these is 3.

Question 17

If the roots of the equation x2+bx+c=0x^2 + bx + c = 0 are -2 and 6, what is the value of bb?

  1. -12
  2. -8
  3. -4 (correct answer)
  4. 4
Explanation: The sum of the roots of a quadratic equation ax2+bx+c=0ax^2+bx+c=0 is −b/a-b/a. In x2+bx+c=0x^2+bx+c=0, a=1a=1, so the sum of the roots is −b-b. The given roots are -2 and 6. Their sum is −2+6=4-2 + 6 = 4. Therefore, −b=4-b = 4, which means b=−4b = -4.

Question 18

The equation 5x2+15x−30=05x^2 + 15x - 30 = 0 has two roots, r1r_1 and r2r_2. What is the value of r1+r2r_1 + r_2?

  1. -6
  2. -3 (correct answer)
  3. 3
  4. 6
Explanation: The sum of the roots of a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0 is given by the formula −b/a-b/a. For the equation 5x2+15x−30=05x^2 + 15x - 30 = 0, we have a=5a=5 and b=15b=15. Therefore, the sum of the roots is −15/5=−3-15/5 = -3.

Question 19

What is the absolute difference between the two solutions of the equation x2−9x+18=0x^2 - 9x + 18 = 0?

  1. 3 (correct answer)
  2. 6
  3. 9
  4. 18
Explanation: First, find the solutions to the quadratic equation. The equation x2−9x+18=0x^2 - 9x + 18 = 0 can be factored by finding two numbers that multiply to 18 and add to -9. These numbers are -6 and -3. So the equation becomes (x−6)(x−3)=0(x-6)(x-3) = 0. The solutions are x=6x=6 and x=3x=3. The absolute difference between these solutions is ∣6−3∣=3|6 - 3| = 3.

Question 20

What is the product of the roots of the equation 2x2−13x−7=02x^2 - 13x - 7 = 0?

  1. -7
  2. -7/2 (correct answer)
  3. 7/2
  4. 13/2
Explanation: The product of the roots of a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0 is given by the formula c/ac/a. In this equation, a=2a=2 and c=−7c=-7. Therefore, the product of the roots is −7/2-7/2.