Question

How to do this question(7)? Pls give explanation as well

Answer 1

$\texttt{\textsf{\large{\underline{Answer}:}}}$

• The value of the expression is 3.

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Let us assume that:

$\sf \implies x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } }$

We can also write it as:

$\sf \implies x = \sqrt{6 + x }$

Squaring both sides, we get:

$\sf \implies {x}^{2} =6 + x$

$\sf \implies {x}^{2} - x - 6 =0$

By splitting the middle term:

$\sf \implies {x}^{2} - 3x + 2x - 6 =0$

$\sf \implies x(x - 3) + 2(x - 3 ) =0$

$\sf \implies (x+ 2)(x - 3 ) =0$

Therefore:

$\implies\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}$

$\sf \implies x = - 2,3$

But x cannot be negative.

$\sf \implies x = 3$

★ So, the value of the expression is 3.

$\texttt{\textsf{\large{\underline{Verification}:}}}$

Given:

$\sf\implies x=3$

We can also write it as:

$\sf\implies x = \sqrt{9}$

$\sf\implies x = \sqrt{6+3}$

$\sf\implies x = \sqrt{6 + \sqrt{9}}$

$\sf\implies x = \sqrt{6 + \sqrt{6+3}}$

$\sf\implies x = \sqrt{6 + \sqrt{6+\sqrt{9}}}$

$\sf\implies x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}$

This pattern will continue.

$\sf\implies x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}$

Hence, our answer is correct. (Verified)