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 factorx2 + 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. Hencex2 + 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 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)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 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) 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 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 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. 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 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 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 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 Maths tests Free Compass Maths tests Practice Free Practice for SAT, ACT Maths tests Free GRE Quantitative for Practice Free AP Calculus Questions (AB and BC) with Answers |
