Question
Solve the following equation by factorisation :
$
\sqrt{x+15}=x+3
$
$
\sqrt{x+15}=x+3
$
✓
Answer
$
\begin{aligned}
& \sqrt{x+15}=x+3 \\
& \text { Squaring on both sides } \\
& x+15=(x+3)^2 \\
& \Rightarrow x^2+6 x+9-x-15=0 \\
& \Rightarrow x^2+5 x-6=0 \\
& \Rightarrow x^2+6 x-x-6=0 \\
& \Rightarrow x(x+6)-1(x+6)=0 \\
& \Rightarrow(x+6)(x-1)=0 \\
& \text { Either } x+6=0 \\
& \text { then } x=-6 \\
& \text { or } \\
& x-1=0, \\
& \text { then } x=1 \\
& \therefore x=-6,1
\end{aligned}
$
Check:
(i) If $x=-6$ then
$
\begin{aligned}
& \text { L.H.S. }=\sqrt{x+15} \\
& =\sqrt{-6+15} \\
& =\sqrt{9} \\
& =3 \\
& \text { R.H.S. }=x+3 \\
& =-6+3 \\
& =-3 \\
& \because \text { L.H.S. } \neq \text { R.H.S. }
\end{aligned}
$
$
\therefore x=-6 \text { is not a root }
$
(ii) If $x =1$, then
$
\begin{aligned}
& \text { L.H.S. }-\sqrt{x+15} \\
& =\sqrt{1+15} \\
& =\sqrt{16} \\
& =4
\end{aligned}
$
$
\begin{aligned}
& \text { R.H.S. }=x+3 \\
& =1+3 \\
& =4
\end{aligned}
$
$\because$ L.H.S. $=$ R.H.S.
$\therefore x=1$ is a root of this equation
Hence $x=1$.
\begin{aligned}
& \sqrt{x+15}=x+3 \\
& \text { Squaring on both sides } \\
& x+15=(x+3)^2 \\
& \Rightarrow x^2+6 x+9-x-15=0 \\
& \Rightarrow x^2+5 x-6=0 \\
& \Rightarrow x^2+6 x-x-6=0 \\
& \Rightarrow x(x+6)-1(x+6)=0 \\
& \Rightarrow(x+6)(x-1)=0 \\
& \text { Either } x+6=0 \\
& \text { then } x=-6 \\
& \text { or } \\
& x-1=0, \\
& \text { then } x=1 \\
& \therefore x=-6,1
\end{aligned}
$
Check:
(i) If $x=-6$ then
$
\begin{aligned}
& \text { L.H.S. }=\sqrt{x+15} \\
& =\sqrt{-6+15} \\
& =\sqrt{9} \\
& =3 \\
& \text { R.H.S. }=x+3 \\
& =-6+3 \\
& =-3 \\
& \because \text { L.H.S. } \neq \text { R.H.S. }
\end{aligned}
$
$
\therefore x=-6 \text { is not a root }
$
(ii) If $x =1$, then
$
\begin{aligned}
& \text { L.H.S. }-\sqrt{x+15} \\
& =\sqrt{1+15} \\
& =\sqrt{16} \\
& =4
\end{aligned}
$
$
\begin{aligned}
& \text { R.H.S. }=x+3 \\
& =1+3 \\
& =4
\end{aligned}
$
$\because$ L.H.S. $=$ R.H.S.
$\therefore x=1$ is a root of this equation
Hence $x=1$.
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