InterviewSolution
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InterviewSolution
| 1. |
If -5 is a root of the quadratic equation 2x |
| Answer» Given that -5 is the root of\xa0{tex}2 x^{2}+p x-15=0{/tex}Put x = -5 in\xa0{tex}2 x^{2}+p x-15=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2(-5)^{2}+p(-5)-15=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}50-5 p-15=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}35 - 5p = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5p = 35\xa0{/tex}{tex}\\therefore{/tex}\xa0{tex}p = 7{/tex}Hence the quadratic equation p {tex}(x^2 + x) + k = 0{/tex} becomes,\xa0{tex}7\\left(x^{2}+x\\right)+k=0{/tex}{tex}7 x^{2}+7 x+k=0{/tex}\xa0Here {tex}a = 7,\\ b = 7\\ and\\ c = k{/tex} Given that this quadratic equation has equal roots\xa0{tex}\\therefore{/tex}\xa0{tex}b^{2}-4 a c=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}7^{2}-4(7)(\\mathrm{k})=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}49 - 28k = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}49 = 28k\xa0{/tex}{tex}\\therefore{/tex}\xa0k = {tex}\\frac{49} {28}{/tex}\xa0=\xa0{tex}\\frac{7} {4}{/tex} | |