Question

$( \sqrt{x} - 1) {}^{2} = 8 - \sqrt{28}$
find the value of x​

Answer 1

First, let's check if the right-hand side is a perfect square of some irrational number.

$\text{ \begin{aligned}\cdots\longrightarrow8-\sqrt{28}&=7-2\sqrt{7}+1\\\\&=(\sqrt{7})^{2}-2\sqrt{7}+1\\\\&=\underline{(\sqrt{7}-1)^{2}}.\end{aligned} }$

We have a difference of two perfect squares, so we can use the polynomial identity.

$\text{ \cdots\longrightarrow\boxed{a^{2}-b^{2}=(a+b)(a-b)} }$

Then,

$\text{ \cdots\longrightarrow\{(\sqrt{x}-1)+(\sqrt{7}-1)\}\{(\sqrt{x}-1)-(\sqrt{7}-1)\}=0 }$

$\text{ \cdots\longrightarrow\underline{(\sqrt{x}+\sqrt{7}-2)(\sqrt{x}-\sqrt{7})=0}. }$

And then,

$\text{ \cdots\longrightarrow\sqrt{x}+\sqrt{7}-2=0 or \sqrt{x}-\sqrt{7}=0 }$

$\text{ \cdots\longrightarrow\underline{\sqrt{x}=2-\sqrt{7}} or \underline{\sqrt{x}-\sqrt{7}=0} .}$

The first equation does not have a positive solution, as the left-hand side and right-hand side have different signs.

$\cdots\longrightarrow x=7.$

So, the required value is,

$\cdots\longrightarrow\boxed{x=7.}$

Answer 2

$\sf\boxed{x=7}$

$\sf\underline \red{ \bold Question:-}$

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

$\sf\underline\pink{ \bold To \: \bold Find:-}$

$\longrightarrow \sf\: to \: find \: the \: value \: of \: x$

$\sf\underline\green { \bold Solution:-}$

It is given that, (√x - 1)² = 8 - √28

⇒(√x - 1)² = 8 - √(2 × 2 × 7)

★ we know, (a ± b)² = a² + b² ± 2ab

so, (√x - 1)² = (√x)² + 1² - 2(1)(√x) = x + 1 - 2√x ]

⇒(x + 1 - 2√x) = 8 - 2√7

⇒x + 1 - 2√x = 8 - 2√7

⇒x - 2√x = 8 - 1 - 2√7

⇒x - 2√x = 7 - 2√7

on comparing both sides, we get

x = 7

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