Question
Solve the following equation by factorization$\sqrt{3 x+4}=x$
✓
Answer
$
\sqrt{3 x+4}=x
$Squaring on both sides
$
\begin{aligned}
& 3 x+4=x^2 \\
& \Rightarrow x^2-3 x-4=0 \\
& \Rightarrow 4 x+x-4=0 \\
& \Rightarrow x(x-4)+1(x-4)=0 \\
& \Rightarrow(x-4)(x+1)=0
\end{aligned}
$
Either $x-4=0$,
then $x=4$
or
$x+1=0$,
then $x =-1$
$\therefore x=4,-1$
Check
(i) If $x =4$, then
L.H.S.
$=\sqrt{3 x+4}$
$=\sqrt{3 \times 4+4}$
$=\sqrt{12+4}$
$=\sqrt{16}$
$=4$
R.H.S.
$= x$
$=4$
$\therefore$ L.H.S. $=$ R.H.S.
Hence $x=4$ is its root
(ii) If $x=-1$, then
L.H.S.
$
\begin{aligned}
& =\sqrt{3 x(-1)+4} \\
& =\sqrt{-3+4} \\
& =\sqrt{1} \\
& =1
\end{aligned}
$
R.H.S.
$= x$
$=-1$
$\because$ L.H.S. $\neq$ R.H.S.
$\therefore x =-1$ is not its root,
Hence x = 4.
\sqrt{3 x+4}=x
$Squaring on both sides
$
\begin{aligned}
& 3 x+4=x^2 \\
& \Rightarrow x^2-3 x-4=0 \\
& \Rightarrow 4 x+x-4=0 \\
& \Rightarrow x(x-4)+1(x-4)=0 \\
& \Rightarrow(x-4)(x+1)=0
\end{aligned}
$
Either $x-4=0$,
then $x=4$
or
$x+1=0$,
then $x =-1$
$\therefore x=4,-1$
Check
(i) If $x =4$, then
L.H.S.
$=\sqrt{3 x+4}$
$=\sqrt{3 \times 4+4}$
$=\sqrt{12+4}$
$=\sqrt{16}$
$=4$
R.H.S.
$= x$
$=4$
$\therefore$ L.H.S. $=$ R.H.S.
Hence $x=4$ is its root
(ii) If $x=-1$, then
L.H.S.
$
\begin{aligned}
& =\sqrt{3 x(-1)+4} \\
& =\sqrt{-3+4} \\
& =\sqrt{1} \\
& =1
\end{aligned}
$
R.H.S.
$= x$
$=-1$
$\because$ L.H.S. $\neq$ R.H.S.
$\therefore x =-1$ is not its root,
Hence x = 4.
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