Search any question & find its solution
Question:
Answered & Verified by Expert
If the equations $x^2+a x+b=0$ and $x^2+b x+a=0(a \neq b)$ have a common root, then $a+b$ is equal to
Options:
Solution:
2693 Upvotes
Verified Answer
The correct answer is:
$-1$
Let $\alpha$ be the common root, then
$$
\begin{array}{rlrl}
& \alpha^2+a \alpha+b & =0 \\
\text { and } & \alpha^2+b \alpha+a & =0 \\
\therefore & & \frac{\alpha^2}{a^2-b^2} & =\frac{\alpha}{b-a}=\frac{1}{b-a} \\
\Rightarrow & & \frac{\alpha}{b-a} & =\frac{1}{b-a} \Rightarrow \alpha=1 \\
\text { Now, } \quad & \frac{\alpha^2}{a^2-b^2} & =\frac{\alpha}{b-a} \\
\Rightarrow & \frac{1}{(a+b)(a-b)} & =\frac{1}{-(a-b)} \\
\Rightarrow & a+b & =-1
\end{array}
$$
$$
\begin{array}{rlrl}
& \alpha^2+a \alpha+b & =0 \\
\text { and } & \alpha^2+b \alpha+a & =0 \\
\therefore & & \frac{\alpha^2}{a^2-b^2} & =\frac{\alpha}{b-a}=\frac{1}{b-a} \\
\Rightarrow & & \frac{\alpha}{b-a} & =\frac{1}{b-a} \Rightarrow \alpha=1 \\
\text { Now, } \quad & \frac{\alpha^2}{a^2-b^2} & =\frac{\alpha}{b-a} \\
\Rightarrow & \frac{1}{(a+b)(a-b)} & =\frac{1}{-(a-b)} \\
\Rightarrow & a+b & =-1
\end{array}
$$
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.