Question

$\sqrt{8 } - \sqrt{x} = \sqrt{x} - \sqrt{4} findx \:$
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Answer 1

Answer:

√8-√x = √x-√4

2√2-√x=√x-2

2√2-2=√x+√x

2(√2-1)=2√x

√x=√2-1

squaring on both sides

x=(√2-1)²

using ( a-b) ²=a ²+b²-2ab

x=(√2)²+(1)-2(√2)(1)

x=2+1-2√2

x=3-2√2

Answer 2

given Question:

$\sqrt{8} - \sqrt{x} = \sqrt{x} - \sqrt{4} \:$

$\sqrt{x} + \sqrt{x} = - \sqrt{8} - \sqrt{4}$

$2 \sqrt{x} = - ( \sqrt{8} + \sqrt{4} )$

squaring both the side

${(2 \sqrt{x} )}^{2} = - {( \sqrt{8} - \sqrt{4} ) }^{2}$

$4x = - (8 + 4 + 2 \sqrt{32} )$

$4x = - 12 - 8 \sqrt{2}$

$x = \frac{ - (12 + 8 \sqrt{2}) }{4}$

hope it help ✔