Quadratic Equations practice questions

Quadratic Equations questions

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1. 1Quadratic equationsStandard

   Which value of x is a solution to x² - 5x + 6 = 0?Which value of x is a solution to x^2 - 5x + 6 = 0?

   • A11
   • B22
   • C44
   • D66
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   Answer: 22

   Factor the expression: (x - 2)(x - 3) = 0. The solutions are 2 and 3, so 2 is the listed solution.Factor the expression: (x - 2)(x - 3) = 0. The solutions are 2 and 3, so 2 is the listed solution.

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2. 2Quadratic equationsStandard

   How many real solutions does x² + 4x + 5 = 0 have?How many real solutions does x^2 + 4x + 5 = 0 have?

   • AInfinitely manyInfinitely many
   • B22
   • C00
   • D11
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   Answer: 00

   The discriminant is 4² - 4(1)(5) = 16 - 20 = -4. A negative discriminant means there are no real solutions.The discriminant is 4^2 - 4(1)(5) = 16 - 20 = -4. A negative discriminant means there are no real solutions.

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3. 3Quadratic equationsStandard

   What is the sum of the solutions to x² - 9x + 14 = 0?What is the sum of the solutions to x^2 - 9x + 14 = 0?

   • A1414
   • B55
   • C99
   • D-9-9
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   Answer: 99

   For ax² + bx + c = 0 the solutions sum to -b/a, which is 9 here. Factoring confirms it: (x - 2)(x - 7) gives 2 and 7.For ax^2 + bx + c = 0 the solutions sum to -b/a, which is 9 here. Factoring confirms it: (x - 2)(x - 7) gives 2 and 7.

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4. 4Quadratic equationsStandard

   What is the product of the solutions to x² - 4x - 21 = 0?What is the product of the solutions to x^2 - 4x - 21 = 0?

   • A-4-4
   • B44
   • C2121
   • D-21-21
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   Answer: -21-21

   For ax² + bx + c = 0 the solutions multiply to c/a, which is -21. Factoring confirms it: (x - 7)(x + 3) gives 7 and -3.For ax^2 + bx + c = 0 the solutions multiply to c/a, which is -21. Factoring confirms it: (x - 7)(x + 3) gives 7 and -3.

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5. 5Quadratic equationsStandard

   What is the positive difference between the solutions to x² - 10x + 21 = 0?What is the positive difference between the solutions to x^2 - 10x + 21 = 0?

   • A1010
   • B44
   • C2121
   • D33
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   Answer: 44

   The expression factors as (x - 3)(x - 7), so the solutions are 3 and 7 and the difference is 4. The other choices are the sum of the solutions, their product, and one solution on its own.The expression factors as (x - 3)(x - 7), so the solutions are 3 and 7 and the difference is 4. The other choices are the sum of the solutions, their product, and one solution on its own.

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6. 6Quadratic equationsStandard

   What are the solutions to x² = 6x?What are the solutions to x^2 = 6x?

   • A6 only6 only
   • B-6 and 6-6 and 6
   • C0 and 60 and 6
   • D0 and -60 and -6
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   Answer: 0 and 60 and 6

   Bring everything to one side: x² - 6x = 0, so x(x - 6) = 0 and x is 0 or 6. Dividing both sides by x is what produces "6 only" — and it silently divides by zero, losing the other solution.Bring everything to one side: x^2 - 6x = 0, so x(x - 6) = 0 and x is 0 or 6. Dividing both sides by x is what produces "6 only" — and it silently divides by zero, losing the other solution.

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7. 7Quadratic equationsStandard

   What are the solutions to x² + 2x - 15 = 0?What are the solutions to x^2 + 2x - 15 = 0?

   • A5 and -35 and -3
   • B5 and 35 and 3
   • C-5 and -3-5 and -3
   • D-5 and 3-5 and 3
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   Answer: -5 and 3-5 and 3

   The expression factors as (x + 5)(x - 3), so the solutions are -5 and 3. The signs of the solutions are the opposite of the signs in the brackets, which is what "5 and -3" gets wrong.The expression factors as (x + 5)(x - 3), so the solutions are -5 and 3. The signs of the solutions are the opposite of the signs in the brackets, which is what "5 and -3" gets wrong.

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8. 8Quadratic equationsStandard

   What are the solutions to x² - 7x + 12 = 0?What are the solutions to x^2 - 7x + 12 = 0?

   • A-3 and -4-3 and -4
   • B3 and 43 and 4
   • C3 and -43 and -4
   • D12 and 112 and 1
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   Answer: 3 and 43 and 4

   The expression factors as (x - 3)(x - 4). The constants must multiply to 12 and add to -7, which -3 and -4 do, giving solutions 3 and 4.The expression factors as (x - 3)(x - 4). The constants must multiply to 12 and add to -7, which -3 and -4 do, giving solutions 3 and 4.

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9. 9Quadratic equationsStandard

   What are the solutions to x² - 11x + 30 = 0?What are the solutions to x^2 - 11x + 30 = 0?

   • A-5 and -6-5 and -6
   • B5 and 65 and 6
   • C5 and -65 and -6
   • D30 and 130 and 1
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   Answer: 5 and 65 and 6

   The expression factors as (x - 5)(x - 6): the two constants multiply to 30 and add to -11. The solutions are the opposites of the signs in the brackets.The expression factors as (x - 5)(x - 6): the two constants multiply to 30 and add to -11. The solutions are the opposites of the signs in the brackets.

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10. 10Quadratic equationsChallenge

    Which expression gives the solutions to 2x² + 3x - 4 = 0?Which expression gives the solutions to 2x^2 + 3x - 4 = 0?

    • A3 ± √414(3 ± √(41))/4
    • B-3 ± √414(-3 ± √(41))/4
    • C-3 ± √412(-3 ± √(41))/2
    • D-3 ± √234(-3 ± √(23))/4
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    Answer: -3 ± √414(-3 ± √(41))/4

    This quadratic does not factor, so apply the quadratic formula, x = -b ± √b² - 4ac2a, with a = 2, b = 3, c = -4. The discriminant is 9 - 4(2)(-4) = 9 + 32 = 41, and 2a = 4.This quadratic does not factor, so apply the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), with a = 2, b = 3, c = -4. The discriminant is 9 - 4(2)(-4) = 9 + 32 = 41, and 2a = 4.

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11. 11Quadratic equationsChallenge

    For what value of c does x² - 6x + c = 0 have exactly one real solution?For what value of c does x^2 - 6x + c = 0 have exactly one real solution?

    • A66
    • B-9-9
    • C99
    • D33
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    Answer: 99

    One real solution means the discriminant is zero: 36 - 4c = 0, so c = 9. The equation is then (x - 3)² = 0.One real solution means the discriminant is zero: 36 - 4c = 0, so c = 9. The equation is then (x - 3)^2 = 0.

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12. 12Quadratic equationsChallenge

    For which positive value of k does x² + kx + 16 = 0 have exactly one real solution?For which positive value of k does x^2 + kx + 16 = 0 have exactly one real solution?

    • A44
    • B88
    • C1616
    • D3232
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    Answer: 88

    One real solution means the discriminant is zero: k² - 4(1)(16) = k² - 64 = 0, so k is 8 or -8, and the positive value is 8. Then the equation is x² + 8x + 16 = (x + 4)², whose only root is -4.One real solution means the discriminant is zero: k^2 - 4(1)(16) = k^2 - 64 = 0, so k is 8 or -8, and the positive value is 8. Then the equation is x^2 + 8x + 16 = (x + 4)^2, whose only root is -4.

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