Question

(√x-1)^2=8-√28 ka solve question
Batao

Answer 1

x-2√x+1=8-2√7
x-2√x = 7-2√7
therefore we can conclude that
x= 7

Answer 2

The value of x is 7

Given:

$\sqrt{(x-1)^{2}}=8-\sqrt{28}$

To find:

The value of x

Solution:

Let, left hand side = L.H.S.

Let, right hand side = R.H.S.

Consider the L.H.S.,  

We know that, $\left(a^{2}-b^{2}=a^{2}+b^{2}-2 a b\right)$

Now, by applying the above formula,  

$\begin{array}{l}{\left(\sqrt{x )^{2}}+(1)^{2}-2 \times(1) \times(\sqrt{x})\right.} \\ {x+1-2 \times \sqrt{x}=8-\sqrt{28}}\end{array}$

Consider the R.H.S.,  

$\begin{array}{l}{8-\sqrt{28}=8-\sqrt{(7 \times 4)}} \\ {8-\sqrt{28}=8-\sqrt{\left(7 \times(2)^{2}\right)}}\end{array}$

Since, $\sqrt{2 )^{2}}=2$ , we get

$8-\sqrt{28}=8-2 \times \sqrt{(7)}$

Hence, $x+1-2 \times \sqrt{x}=8-2 \times \sqrt{(7)}$

$\begin{array}{l}{x-2 \times \sqrt{x}=7-2 \times \sqrt{(7)}} \\ {\sqrt{x}(\sqrt{x}-2)=\sqrt{7}(\sqrt{7}-2)}\end{array}$

Therefore, by considering the L.H.S and R.H.S, it is understood that $x=7$