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The roots
$\begin{aligned}
& (x-a)(x-a-1)+(x-a-1)(x-a-2) \\
& +(x-a)(x-a-2)=0, a \in R \text { are always }
\end{aligned}$
Options:
$\begin{aligned}
& (x-a)(x-a-1)+(x-a-1)(x-a-2) \\
& +(x-a)(x-a-2)=0, a \in R \text { are always }
\end{aligned}$
Solution:
2135 Upvotes
Verified Answer
The correct answer is:
real and distinct
Given,
$\begin{aligned}
& (x-a)(x-a-1)+(x-a-1)(x-a-2) \\
& +(x-a)(x-a-2)=0
\end{aligned}$
Let $x-a=t$, then
$t(t-1)+(t-1)(t-2)+t(t-2)=0$
$\begin{array}{lc}\Rightarrow & t^2-t+t^2-3 t+2+t^2-2 t=0 \\ \Rightarrow & 3 t^2-6 t+2=0 \\ \Rightarrow & t=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)} \\ \Rightarrow & x-a=\frac{3 \pm \sqrt{3}}{3} \\ \Rightarrow & x=a+\frac{3 \pm \sqrt{3}}{3}\end{array}$
Hence, x is real and distinct.
$\begin{aligned}
& (x-a)(x-a-1)+(x-a-1)(x-a-2) \\
& +(x-a)(x-a-2)=0
\end{aligned}$
Let $x-a=t$, then
$t(t-1)+(t-1)(t-2)+t(t-2)=0$
$\begin{array}{lc}\Rightarrow & t^2-t+t^2-3 t+2+t^2-2 t=0 \\ \Rightarrow & 3 t^2-6 t+2=0 \\ \Rightarrow & t=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)} \\ \Rightarrow & x-a=\frac{3 \pm \sqrt{3}}{3} \\ \Rightarrow & x=a+\frac{3 \pm \sqrt{3}}{3}\end{array}$
Hence, x is real and distinct.
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