Search any question & find its solution
Question:
Answered & Verified by Expert
The roots of
$\begin{aligned}
(x-a)(x-a-1)+(x-a-1)(x-a-2) & \\
&+(x-a)(x-a-2)=0
\end{aligned}$
$a \in R$ are always
Options:
$\begin{aligned}
(x-a)(x-a-1)+(x-a-1)(x-a-2) & \\
&+(x-a)(x-a-2)=0
\end{aligned}$
$a \in R$ are always
Solution:
1710 Upvotes
Verified Answer
The correct answer is:
real and distinct
$\begin{aligned}(x-a)(x-a-1)+&(x-a-1)(x-a-2) \\ &+(x-a)(x-a-2)=0 \end{aligned}$
Let $x-a=t$, then
$$
\begin{array}{l}
t(t-1)+(t-1)(t-2)+t(t-2)=0 \\
\Rightarrow t^{2}-t+t^{2}-3 t+2+t^{3}-2 t=0 \\
\Rightarrow 3 t^{2}-6 t+2=0 \\
\Rightarrow t=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)} \\
\Rightarrow x-a=\frac{3 \pm \sqrt{3}}{3} \\
\Rightarrow x=a+\frac{3 \pm \sqrt{3}}{3}
\end{array}
$$
Hence, $x$ is real and distinct.
Let $x-a=t$, then
$$
\begin{array}{l}
t(t-1)+(t-1)(t-2)+t(t-2)=0 \\
\Rightarrow t^{2}-t+t^{2}-3 t+2+t^{3}-2 t=0 \\
\Rightarrow 3 t^{2}-6 t+2=0 \\
\Rightarrow t=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)} \\
\Rightarrow x-a=\frac{3 \pm \sqrt{3}}{3} \\
\Rightarrow x=a+\frac{3 \pm \sqrt{3}}{3}
\end{array}
$$
Hence, $x$ is real and distinct.
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.