2 votes 2 votes Let $f(x) = ax^2 + bx +c$, where $a$, $b$ and $c$ are certain constants and $a \neq 0$. It is known that $f(5) = -3 f(2) $ and that 3 is a root of $f(x)=0$. What is the other root of $f(x)=0?$ $-7$ $-4$ $2$ $6$ cannot be determined Quantitative Aptitude cat2008 quantitative-aptitude quadratic-equations + – go_editor asked Nov 28, 2015 • retagged Mar 28, 2022 by Lakshman Bhaiya go_editor 13.9k points 908 views answer comment Share See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes 1)3 is root So we get 9a+3b+c=0 Acc to second condition f(5)=-3f(2) 25a+5b+c=-3(4a+2b+c) 37a+11b+4c=0 Solving this two eqn we get a=b and c=-12a ax^2+bx+c=ax^2+ax-12a =a(x^2+x-12) =a(x+4)(x-3) So other root is - 4 Pooja Palod answered Dec 5, 2015 • selected Jan 6, 2017 by Vijay Thakur Pooja Palod 1.8k points comment Share See all 0 reply Please log in or register to add a comment.