Question

x = √3/2 then find the value of
√(1 + x ) + √(1 -x ) =?

Answer 1

let √(1 + x ) + √(1 -x ) = θ
squaring on both sides

1+x+1-x+{2 [√1+x*1-x] } =θ²

2+2√(1²-x²)=θ²

2+2√(1-(√3/2)²)=θ²

2+2√(4-3)/4=θ²

2+2√(1/4)=θ²

2+2*1/2=θ²

2+1=θ²

3=θ²

√3=θ

θ=1.732


hope helped!!

Answer 2

Given : x = $\dfrac{\sqrt{3}}{2}$

Given equation : $\sqrt{(1+x)}+\sqrt{(1-x)}$

⇒$\implies\sqrt{\bigg( \sqrt{(1+x)} +\sqrt{(1-x)}\bigg)^{2}} \\ \\\\ \\ \implies\sqrt{ \bigg(\sqrt{1+x}\bigg)^2 + \bigg(\sqrt{ 1 - x } \bigg)^2 +2 \bigg(\sqrt{1 +x)(1-x) }\bigg) }$

$\implies \sqrt{1+ 1 + x - x +2 \bigg(\sqrt{1 - x^2 }\bigg) } }$

$\implies \sqrt{ 2 + 2 \bigg(\sqrt{1-x^2 }\bigg) }$

Now, substituting the value of x which is given in the question,

$\implies\sqrt{2 + 2\sqrt{\bigg[1- \bigg(\dfrac{\sqrt{3}}{2} \bigg)^2\bigg]}}\\ \\ \\ \\ \implies \sqrt{2 + 2 \sqrt{\bigg( \dfrac{4-3}{4} \bigg)}} \\ \\ \\ \\ \implies \sqrt{2 +\bigg( 2 \times \dfrac{1}{2} \bigg)} \\ \\ \\ \\ \implies \sqrt{2 + 1 }$

Therefore the value of √( x + 1 ) + √( x - 1 ) is √3 .