Question

the square root of X+√x^2-y^2 is given by-​

Answer 1

Given : $X+\sqrt{x^2-y^2}$

To find : The squire root of $X+\sqrt{x^2-y^2}$.

Step-by-step explanation:

At first let squire root of $X+\sqrt{x^2-y^2}$ is $I$, that is,

$I=\sqrt{ X+\sqrt{x^2-y^2}}\hfill (1)$

Since both terms x and y in the root $\sqrt{x^2-y^2}$ are perfect squires we can apply difference of squire formula to factor it.

That is,

$x^2-y^2=(x+y)(x-y)$

Substitute this value in (1) we get,

$\therefore I=\sqrt{X+\sqrt{(x+y)(x-y)}}$

which is the required solution.

Answer 2

Step-by-step explanation:

Let us consider

P = x+√x^2-y^2

P = x+√(x+y)(x-y)

Multiplying and Dividing the whole equation by 2

P = 1/2[2x+2√(x+y)(x-y)]

P = 1/2[2x+y-y+2√(x+y)(x-y)]

P = 1/2[(x+y) + (x-y) + 2√(x+y)(x-y)]

The equation in the square bracket is (a+b)^2 formula

Therefore

P = 1/2[(√x+y + √x-y)^2]

Taking the root of both sides

√P = 1/√2(√x+y + √x-y)