# 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.
## Solution to Question 1Write the given equation with right hand side equal to 0 and factorx ^{2} + 5x + 6 = 0
(x + 2)(x + 3) Solve for x x + 2 = 0 or x + 3 = 0 Solutions: x = - 2 and x = - 3 ## Solution to Question 2Since x = 1 and x = -2 are solutions to the equation, they must satisfy the equation2(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 ## Solution to Question 3Rewrite 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 ## Solution to Question 4Subtract 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 ## Solution to Question 5For the given equation to have a solution, the left hand side must be a square. Hencex ^{2} + k x + 4 = (x + 2)^{2}Expand the right hand side x ^{2} + k x + 4 = x^{2} + 4x + 4
Comparing the left hand side and the right hand sides, the value of k is 4 k = 4 ## Solution to Question 6x = 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)x ^{2} = 16
Substitute m by 2 in the given equation and solve it. 4 x ^{2} = 16
x ^{2} = 4
x = ~+mn~ 2 The second solution is x = - 2 ## Solution to Question 7x = - 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) x ^{2} + 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 ## Solution to Question 8Since 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 x ^{2} - x - 2 = 0
If we now compare the given equation x ^{2} + bx + c = 0 with the obtained equation x^{2} - x - 2 = 0, we can identify constants b and c by their values as followsb = -1 and c = -2. ## Solution to Question 9The 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 X ^{2} + 2 = 11 x
5 X ^{2} - 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 ## Solution to Question 10The 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 x ^{2} + 3 = 10x
Write the above equation with right hand term equal to 0, factor and solve. 3 x ^{2} - 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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