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. 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 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 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 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 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 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 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 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. 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 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 More Math Practice Tests Free Practice for GAMT Math tests Free Compass Math tests Practice Free Practice for SAT, ACT Math tests Free GRE Quantitative for Practice Free AP Calculus Questions (AB and BC) with Answers