GMAT Problems on Quadratic Equations
with Solutions and Explanations Sample 3

Solutions and detailed explanations to quadratic equations problems in sample 3 similar to the problems in the GMAT test.

  1. Solve the equation x2 + 5x = - 6 for x.

    Solution

    Write the given equation with right hand side equal to 0 and factor

    x2 + 5x + 6 = 0

    (x + 2)(x + 3)

    Solve for x

    x + 2 = 0 or x + 3 = 0

    Solutions: x = - 2 and x = - 3

  2. Find the values of the constants k and m so that the equation 2 x2 + kx - m = 0 has solutions at x = 1 and x = -2.

    Solution

    Since x = 1 and x = -2 are solutions to the equation, they must satisfy the equation

    2(1)2 + k(1) - m = 0 and 2(-2)2 + k(-2) - m = 0

    Simplify and rewrite as a system of two equations in two unknowns k and m.

    k - m = - 2 and -2k - m = -8

    Subtract the left and right hand terms of the two eqautions to obtain an equivalent equation.

    (k - m) - (-2k - m) = (-2) - (-8)

    Simplify.

    3k = 6

    Solve for k.

    k = 2 and

    Use equation k - m = - 2 to find m.

    m = k + 2 = 4

    The values of k and m are

    k = 2 and m = 4

  3. Solve the equation (x - 1)(x + 3) = (1 - x) for x.

    Solution

    Rewrite equation with the right hand side equal to 0.

    (x - 1)(x + 3) - (1 - x) = 0

    Rewrite as

    (x - 1)(x + 3) + (x - 1) = 0

    Factor and solve

    (x - 1) [ (x + 3) + 1 ] = 0

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

    x - 1 = 0 or x + 4 = 0

    Solutions: x = 1 and x = -4

  4. Solve the equation 2 - (x - 2)2 = - 18 for x.

    Solution

    Subtract 2 from both sides of the equation.

    - (x - 2)2 = - 20

    Multiply both sides of the equation by -1.

    (x - 2)2 = 20

    Solve by extracting the square root.

    x - 2 = ~+mn~ √20 = ~+mn~ 2 √5

    solutions: x = 2 + 2 √5 and x = 2 - 2 √5


  5. Find a positive value of the constant k so that the equation x2 + k x + 4 = 0 has only one solution.

    Solution

    For the given equation to have a solution, the left hand side must be a square. Hence

    x2 + k x + 4 = (x + 2)2

    Expand the right hand side

    x2 + k x + 4 = x2 + 4x + 4

    Comparing the left hand side and the right hand sides, the value of k is 4

    k = 4

  6. If x = 2 is a solution to the equation (m + 2)x2 = 16, where m is a constant, then what is the second solution of this equation?

    Solution

    x = 2 is a solution and therefore satisfies the equation. Hence

    (m + 2)(2)2 = 16

    Solve for m

    4(m + 2) = 16

    m + 2 = 4

    m = 2

    Substitute m by 2 in the given equation and solve it.

    (2 + 2)x2 = 16

    Substitute m by 2 in the given equation and solve it.

    4 x2 = 16

    x2 = 4

    x = ~+mn~ 2

    The second solution is

    x = - 2

  7. If x = - 2 is a solution to the equation (1/2) a x2 + 3x - 4 = 0, where a is a constatnt, then what is the second solution of this equation?

    Solution

    x = - 2 is a solution and therefore satisfies the given equation. Hence

    (1/2) a (-2) 2 + 3x - 4 = 0

    Simplify and solve for a

    (1/2) a (-2) 2 + 3(-2) - 4 = 0 2 a - 6 - 4 = 0

    a = 10 / 2 = 5

    We now substitute a by 5 in the given equation and then factor knowing that x + 2 is a factor since x = - 2 is a solution.

    (5/2) x2 + 3x - 4 = 0

    (x + 2)( (5/2)x - 2 ) = 0

    The second solution is found by setting the second factor equal to 0 and solve

    (5/2)x - 2 = 0

    x = 4 / 5

  8. Find the values of the constants b and c so the equation x2 + bx + c = 0 has solutions at x = 2 and x = -1.

    Solution

    Since the equation has solutions at x = 2 and x = - 1, it can be written in factored form as follows

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

    Expand the left hand side

    x2 - x - 2 = 0

    If we now compare the given equation x2 + bx + c = 0 with the obtained equation x2 - x - 2 = 0, we can identify constants b and c by their values as follows

    b = -1 and c = -2.

  9. X and Y are two real numbers such that X + Y = 11 / 5, X Y = 2 / 5 and X is greater than 1. Find X and Y.

    Solution

    The second equation X Y = 2 / 5 may written as follows

    Y = 2 / (5 X)

    Substitute Y by 2 / (5 X) in the equation X + Y = 11 / 5

    X + 2 / (5 X) = 11 / 5

    Multiply all terms of the above equation and simplify

    5 X2 + 2 = 11 x

    5 X2 - 11 X + 2 = 0

    Factor left hand side and solve

    (5X - 1)(X - 2) =

    5X - 1 = 0 , X = 1 / 5

    X - 2 = 0 , X = 2

    We now calculate the value of Y using the equation X + Y = 11 / 5

    For X = 1 / 5, Y = 11 / 5 - X = 11 / 5 - 1 / 5 = 10 / 5 = 2

    For X = 2, Y = 11 / 5 - X = 11 / 5 - 2 = 11 / 5 - 10 / 5 = 1 / 5

    The two numbers are X = 2 and Y = 1 / 5

  10. The sum of a number x, such that x is greater than 1, and its reciprocal is equal to 10 / 3. Find x.

    Solution

    The reciprocal of x is 1 / x. The sum of x and 1 / x is equal to 10 / 3 is translated mathematically as follows.

    x + 1 / x = 10 / 3

    Multiply all terms in the equation by 3 x and simplify.

    3 x2 + 3 = 10x

    Write the above equation with right hand term equal to 0, factor and solve.

    3 x2 - 10 x + 3 = 0

    (3x - 1)(x - 3) = 0

    Two solutions to the equation

    x = 1 / 3 and x = 3

    We are looking for the solution that is greater than 1. Hence the solution to the problem is

    x = 3



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