Hard SAT and ACT Math Problem with Solution Quadratic Equations Today I would like to provide a solution to yesterday’s hard math problem involving quadratic equations. Here is the problem once again, followed by a solution. Level 5 Quadratic Equations Which of the following is a quadratic equation that has –5/7 as its only solution? A. 49x2 – 70x + 25 = 0 B. 49x2 + 70x + 25 = 0 C. 49x2 + 35x + 25 = 0 D. 49x2 + 25 = 0 E. 49x2 – 25 = 0 * If -5/7 is the only solution, then x + 5/7 is the only factor (by the factor theorem). So we have (x + 5/7)(x + 5/7) = 0 x2 + 5/7 x + 5/7 x + (5/7)2 = 0 x2 + 10/7 x + 25/49 = 0 49x2 + 70x + 25 = 0 This is choice B. Notes: (1) The factor theorem says that r is a root of the polynomial p(x) if and only if x – r is a factor of the polynomial. (2) In factored form, a quadratic equation always has two factors. Since x + 5/7 is the only factor in this problem, it must appear twice. (3) In going from the first equation to the second equation, we multiplied the two polynomials (x + 5/7 times itself). To do this you can use FOIL or any other method you like for multiplying polynomials. (4) In going from the second equation to the third equation, we simply added 5/7 x + 5/7 x = 10/7 x. (5) In going from the third equation to the last equation, we multiplied both sides of the equation by the least common denominator, which is 49. More precisely, on the left hand side we have 49(x2 +10/7 x + 25/49) = 49x2 + 49(10/7 x) + 49(25/49) = 49x2 +7(10x) + 25 = 49x2 +70x + 25. More SAT and ACT Math Problems with Explanations If you are preparing for the SAT or ACT, you may want to take a look at the Get 800 collection of test prep books. And if you liked this article, please share it with your Facebook friends: Comments comments