Question

Solve square root of x + square root of x+2 = 2

Answer 1

The value of x is $\frac{1}{4}$

Step-by-step explanation:

Given:

$\sqrt{x} + \sqrt{x + 2} = 2$

Rearranging the radical on left side,

$\sqrt{x + 2} = 2 - \sqrt{x}$

Power on both sides,

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

Simplifying the left,

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

For the RHS equation, use the property of (a-b)² = (a²-2ab+b²),

$= > x + 2 = {2}^{2} - (2 \times 2 \sqrt{x}) + ( \sqrt{x})^{2}$

Now calculating its powers,

$= > x + 2 = 4 - 4\sqrt{x} + x$

Now sending -4√x to the LHS (left side), its sign becomes plus (+),

$= > 4 \sqrt{x} = 4 + x - x - 2$

Now the +x and -x will be cancelled,

$= > 4 \sqrt{x} = 4 - 2$

$= > 4 \sqrt{x} = 2$

Bringing 4 to the right side, it becomes the denominator,

$= > \sqrt{x} = \frac{2}{4}$

$= > \sqrt{x} = \frac{1}{4}$

Now powering both sides,

$= > ( \sqrt{x})^{2} = ( \frac{1}{2})^{2}$

$= > x = \frac{1}{4}$