Free Compass Math Practice Questions on Quadratic Equations
with Solutions and Explanations - Sample 6



Solutions with detailed explanations to compass math test practice questions in sample 6.

  1. What are the two solutions to the quadratic equations 2 x2 + 3x - 2 = 0?

    Solution

    Factor the left hand of the given equation

    2 x2 + 3x - 2 = 0

    Factor the left hand of the given equation

    (2x - 1)(x + 2) = 0

    Solve

    2x - 1 = 0 , x = 1/2

    x + 2 = 0 , x = - 2

    Solution set

    {-2 , 1/2}

  2. What is the sum of the two solutions to the quadratic equation (x + 4)(x - 3) = 7

    Solution

    The sum of the solutions of a quadratic equation of the form a x2 + bx + c = 0 is given by

    - b / a

    Let us write the given equation in standard form

    (x + 4)(x - 3) = 7

    x2 + x - 12 = 7

    x2 + x - 19 = 0

    The sum of the solutions of the the given equation is

    - b / a = - 1 / 1 = -1

  3. What is the product of the two solutions to the quadratic equation (x - 2)(x - 6) = -3

    Solution

    The product of the solutions of a quadratic equation of the form a x2 + bx + c = 0 is given by

    c / a

    Let us write the given equation in standard form

    (x - 2)(x - 6) = -3

    x2 - 8x + 12 = - 3

    x2 - 8x + 15 = 0

    The product of solutions of the given equation is

    c / a = 15 / 1 = 15


  4. Find all values of m for which the quadratic equation x2 + 2 x - 2 m = 0 have no real solutions.

    Solution

    The type and number of solutions is determined by the sign of the discriminant of the quadratic equation. The discriminant of the given equation is equal to

    22 - 4(1)(-2m) = 4 + 8m

    A quadratic equation has no solutions if it discriminant is negative. Hence

    4 + 8m < 0

    Solve the above inequality to obtain

    m < -1/2

    The given quadratic equation has no real solutions for all values of m such that m < -1/2

  5. Find all values of m for which the quadratic equation 2x2 + 3 x - m + 2 = 0 have two distinct real solutions.

    Solution

    The discriminant of the given equation is equal to

    32 - 4(2)(- m + 2) = 9 + 8m - 16 = -7 + 8m

    A quadratic equation has two distinct real solutions if it discriminant is positive. Hence

    -7 + 8m > 0

    Solve the above inequality to obtain

    m > 7 / 8

    The given quadratic equation has two distinct real solutions for all values of m such that m > 7 / 8

  6. Which of these quadratic equations has two real solutions greater than zero?

    A) x2 + x = 0
    B) 2 x2 - 10 x = 28
    C) - x2 + 4 x + 5 = 0
    D) -3 x2 - 9 = - 12 x
    E) -3 x2 - 6 x + 24 = 0

    Solution

    For the real solutions of quadratic function to be greater than zero, both the sum and the product have to be greater than zero. Let us find the sum (given by -b/a) and product (given by c/a) of the solutions of each of the given equations

    A) x2 + x = 0 , sum = -b/a = -1/1 = - 1, product = c/a = 0/1 = 0
    B) 2 x2 - 10 x - 28 = 0 , sum = -b/a = -(-10)/2 = 5, product = c/a = -28/2 = -14
    C) - x2 + 4 x + 5 = 0 , sum = -b/a = -4/-1 = 4, product = c/a = 5/-1 = -5
    D) -3 x2 + 12 x - 9 = 0 , sum = -b/a = -12/-3 = 4, product = c/a = -9/-3 = 3
    E) -3 x2 - 6 x + 24 = 0 , sum = -b/a = -(-6)/-3 = - 2, product = c/a = 24/-3 = -8

    The equation in D) has both the sum of the solutions and their product positive and therefore its two solutions are both greater zero.


  7. Which of these quadratic equations has two real solutions whose product is greater than zero?

    A) - x2 - 2x = - 8
    B) x2 + 9 x = - 18
    C) - x2 = - 6 + x
    D) x2 = 4 x
    E) x2 - 3 x = 4

    Solution

    Find the product of each of the given equations

    A) - x2 - 2x + 8 = 0 , product = c / a = 8 / -1 = - 8
    B) x2 + 9 x + 18 = 0 , product = c / a = 18 / 1 = 18
    C) - x2 - x + 6 = 0 , product = c / a = 6 / - 1 = -6
    D) x2 - 4 x = 0 , product = c / a = 0 / 1 = 0
    E) x2 - 3 x - 4 = 0 , product = c / a = - 4 / 1 = - 4

    The equation in B) has the product of its solutions positive.


  8. b and c in the quadratic equation x2 + b x + c = 0 are real numbers. Find b and c so that the given equation has two solutions x = -1/4 and x = 1/2.

    Solution

    A quadratic equation with solutions -1/4 and 1/2 and leading coefficient 1 is written as

    (x - (-1/4))(x - 1/2) = 0

    (x + 1/4)(x - 1/2) = 0

    Expand and group like terms

    x2 - (1/2)x + (1/4)x - 1/8 = 0

    x2 - (1/4)x - 1/8 = 0

    Identify b and c

    b = -1/4 , c = -1/8

  9. b and c in the quadratic equation - x2 + bx + c = 0 are real numbers. Find b and c so that the given equation has two solutions whose sum is equal to 6 and whose product is equal to 8.

    Solution

    Sum and product are given by

    sum = - b / -1 = b = 6

    product = c / - 1 = - c = 8 , hence c = -8

    b and c are given by

    b = 6 and c = -8
  10. Which of these pairs of quadratic equations have the same solutions?(equivalent equations)

    A) x2 - 1 = 0 and x2 = - 1
    B) - x2 + x = - 6 and x2 - 2x = 3
    C) x2 - 5x + 6 = 0 and - x2 - 5x - 6 = 0
    D) x2 = 2x and x2 + 2x = 0
    E) x2 + x - 2 = 0 and - 2 x2 -2x + 4 = 0

    Solution

    When written in standard forms with positive leading coefficient, the pairs of given equations become

    A) x2 - 1 = 0 , x2 + 1 = 0
    B) x2 - x - 6 = 0 , x2 - 2x - 3 = 0 (all terms of the first equation have been multiplied by -1)
    C) x2 - 5x + 6 = 0 , x2 + 5x + 6 = 0 (all terms of the second equation have been multiplied by -1)
    D) x2 - 2x = 0 , x2 + 2x = 0
    E) x2 + x - 2 = 0 , x2 + x - 2 = 0 (all terms of the second equations have been divided by -2)

    The two equations in E) has the same solutions

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