Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)$ is a quadratic expression such that $f(1)+f$ (2) $=0$, and $-1$ is a root of $f(x)=0$, then the other root of $f(x)=0$ is
Options:
Solution:
1063 Upvotes
Verified Answer
The correct answer is:
$\frac{8}{5}$
$\frac{8}{5}$
If $a$ and $-1$ are the roots of the polynomial, then we get
$$
\begin{aligned}
& f(x)=x^2+(1-a) x-a . \\
\therefore \quad & f(1)=2-2 a \\
\text { and } & f(2)=6-3 a \\
\text { As, } & f(1)+f(2)=0 \\
\Rightarrow & 2-2 a+6-3 a=0 \Rightarrow a=\frac{8}{5}
\end{aligned}
$$
Therefore, the other root is $\frac{8}{5}$
$$
\begin{aligned}
& f(x)=x^2+(1-a) x-a . \\
\therefore \quad & f(1)=2-2 a \\
\text { and } & f(2)=6-3 a \\
\text { As, } & f(1)+f(2)=0 \\
\Rightarrow & 2-2 a+6-3 a=0 \Rightarrow a=\frac{8}{5}
\end{aligned}
$$
Therefore, the other root is $\frac{8}{5}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.