The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots \ldots .}}}$ is equal
The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots \ldots .}}}$ is equal
$7$
$-6$
$5$
$4$
Solution
Let $y=\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$
$
\Rightarrow \quad y=\sqrt{42+y}
$
On squaring both sides, we get
$
\begin{aligned}
& y^2=42+y \\
& \Rightarrow \quad y^2-y-42=0 \\
&
\end{aligned}
$
$
\begin{array}{rlrl}
\Rightarrow & & (y-7)(y+6) & =0 \\
\Rightarrow & y =7,-6
\end{array}
$
Since, $y=-6$ is not satisfied the given equation.
$\therefore$ The required solution is $y=7$