Saved Bookmarks
| 1. |
21 Find the value of"K, for which x3 is a solution of the quadratic equation,(K + 2)x2-Kx + 6 = 0.Also, find the other root of the equation. |
|
Answer» x = 3 is solution of p(x) = (k+2)x² - kx + 6 then,p(3) = (k+2)×3² - 3k + 6 = 0 9(k+2) - 3k + 6 = 0 9k + 18 - 3k + 6 = 0 6k = -24 k = -24/6 k = -4 So, equation is (-4+2)x² -(-4)x + 6 = 0 -2x² + 4x + 6 = 0 multiply by (-1) 2x² - 4x - 6 = 0 2x² +2x -6x - 6 = 0 2x(x + 2) -6(x + 1) = 0 (2x -6) (x +1) = 0 So, x = 3 and x = -1 but , answer is -1 (written in my book) Our steps are right. I guess it must be printing mistake hints are also given, can I give it to you Sorry for late reply. Yes x = -1, I was not getting your dobut first. check the this solution x = 3 is solution of p(x) = (k+2)x² - kx + 6 then,p(3) = (k+2)×3² - 3k + 6 = 0 9(k+2) - 3k + 6 = 0 9k + 18 - 3k + 6 = 0 6k = -24 k = -24/6 k = -4 So, equation is (-4+2)x² -(-4)x + 6 = 0 -2x² + 4x + 6 = 0 multiply by (-1) 2x² - 4x - 6 = 0 2x² +2x -6x - 6 = 0 2x(x + 2) -6(x + 1) = 0 (2x -6) (x +1) = 0 So, x = 3 and x = -1 in last second step there will be 1 sorry can I ask you one question .. where do you live? |
|