Question
Solve the following equation by factorisation :
$
\sqrt{3 x^2-2 x-1}=2 x-2
$
$
\sqrt{3 x^2-2 x-1}=2 x-2
$
✓
Answer
$
\sqrt{3 x^2-2 x-1}=2 x-2
$
Squaring both sides
$
\begin{aligned}
& 3 x^2-2 x-1=(2 x-2)^2 \\
& \Rightarrow 3 x^2-2 x-1=4 x^2-8 x+4 \\
& \Rightarrow 4 x^2-8 x+4-3 x^2+2 x+1=0 \\
& \Rightarrow x^2-6 x+5=0 \\
& \Rightarrow x^2-5 x-x+5=0 \\
& \Rightarrow x(x-5)-1(x-5)=0 \\
& \Rightarrow(x-5)(x-1)=0
\end{aligned}
$
Either $x-5=0$,
$
\text { then } x=5
$
or
$
x-1=0
$
then $x=1$
Check:
$
\begin{aligned}
& \text { (i) If } x=5 \text {, then } \\
& \text { L.H.S. }=\sqrt{3 x^2-2 x-1} \\
& =\sqrt{3 \times(5)^2-2 \times 5-1} \\
& =\sqrt{3 \times 25-10-1} \\
& =\sqrt{75-10-1} \\
& =\sqrt{64} \\
& =8 \\
& \text { R.H.S. }=2 x-2 \\
& =2 \times 5-2 \\
& =10-2 \\
& =8 \\
& \because \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$\therefore x=5$ is a root
$
\begin{aligned}
& \text { (ii) If } x =1 \text {, then } \\
& \text { L.H.S. }=\sqrt{3 x^2-2 x-1} \\
& =\sqrt{3(1)^2-2(1)-1} \\
& =\sqrt{3 \times 1-2-1} \\
& =\sqrt{3-2-1} \\
& =0 \\
& \text { R.H.S. }=2 x-2 \\
& =2 \times 1-2 \\
& =2-2 \\
& =0 \\
& \because \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$\therefore x =1$ is also its root
Hence $x=5,1$.
\sqrt{3 x^2-2 x-1}=2 x-2
$
Squaring both sides
$
\begin{aligned}
& 3 x^2-2 x-1=(2 x-2)^2 \\
& \Rightarrow 3 x^2-2 x-1=4 x^2-8 x+4 \\
& \Rightarrow 4 x^2-8 x+4-3 x^2+2 x+1=0 \\
& \Rightarrow x^2-6 x+5=0 \\
& \Rightarrow x^2-5 x-x+5=0 \\
& \Rightarrow x(x-5)-1(x-5)=0 \\
& \Rightarrow(x-5)(x-1)=0
\end{aligned}
$
Either $x-5=0$,
$
\text { then } x=5
$
or
$
x-1=0
$
then $x=1$
Check:
$
\begin{aligned}
& \text { (i) If } x=5 \text {, then } \\
& \text { L.H.S. }=\sqrt{3 x^2-2 x-1} \\
& =\sqrt{3 \times(5)^2-2 \times 5-1} \\
& =\sqrt{3 \times 25-10-1} \\
& =\sqrt{75-10-1} \\
& =\sqrt{64} \\
& =8 \\
& \text { R.H.S. }=2 x-2 \\
& =2 \times 5-2 \\
& =10-2 \\
& =8 \\
& \because \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$\therefore x=5$ is a root
$
\begin{aligned}
& \text { (ii) If } x =1 \text {, then } \\
& \text { L.H.S. }=\sqrt{3 x^2-2 x-1} \\
& =\sqrt{3(1)^2-2(1)-1} \\
& =\sqrt{3 \times 1-2-1} \\
& =\sqrt{3-2-1} \\
& =0 \\
& \text { R.H.S. }=2 x-2 \\
& =2 \times 1-2 \\
& =2-2 \\
& =0 \\
& \because \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$\therefore x =1$ is also its root
Hence $x=5,1$.
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