Question

if x= √3/2 find √1+x - √1-x
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Answer 1

ANSWER:

Given:

• x = √3/2

To Find:

• Value of √(1+x) - √(1-x)

Solution:

We need to find that,

$\implies \sqrt{1+x}-\sqrt{1-x}$

Now, we will first take square of the above expression, and then at last, we will take square root.

So on squaring the expression,

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

We know that,

$\hookrightarrow (a-b)^2=a^2+b^2-2ab$

So,

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

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

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

We know that,

$\hookrightarrow (a-b)(a+b)=a^2-b^2$

So,

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

$\implies1+1-2(\sqrt{1-x^2})$

Substituting the value of x,

$\implies2-2\left(\sqrt{1-\left(\dfrac{\sqrt3}{2}\right)^2}\right)$

So,

$\implies2-2\left(\sqrt{1-\dfrac{3}{4}}\right)$

Taking LCM,

$\implies2-2\left(\sqrt{\dfrac{4-3}{4}}\right)$

$\implies2-2\left(\sqrt{\dfrac{1}{4}}\right)$

So,

$\implies2-2\!\!\!/\left(\dfrac{1}{2\!\!\!/}\right)$

Hence,

$\implies2-1$

$\implies1$

Now, we will take square root of the above result,

$\implies\sqrt{1}$

Hence,

$\bf\implies\pm1$

Therefore,

$\implies\bf{\sqrt{1+x}-\sqrt{1-x} = \pm1}\:\:\:(for\:x=\frac{\sqrt3}{2})$