Question

How do i solve for this help please \sqrt(x+5) = \sqrt(x)+1

Answer 1

Answer:

x=4

Step-by-step explanation:

sqrt(x+5) = sqrt(x)+1

Square each side

(sqrt(x+5))^2 = (sqrt(x)+1)^2

x+5 = (sqrt(x)+1)^2

Foil

x+5 = (sqrt(x)) ^2 + sqrt(x) + sqrt(x) + 1

x+5 = x + 2 sqrt(x) + 1

Subtract x from each side

5 = 2 sqrt(x) + 1

Subtract 1 from each sdie

4 = 2 sqrt(x)

Square each side

4^2 = (2 sqrt(x))^2

16 = 4 x

Divide by 4

16/4 = 4x/4

4 =x

Check to see if it is extraneous

sqrt(4+5) = sqrt(4)+1

sqrt(9) = sqrt(4) +1

3 = 2+1

3=3

It is a valid solution

Answer 2

Answer:

$\boxed{x=4}$

Step-by-step explanation:

$\sqrt{x+5} = \sqrt{x}+1$

Take the square on both sides.

$x+5=( \sqrt{x}+1)^2$

Expand brackets.

$x+5=( \sqrt{x}+1) ( \sqrt{x}+1)$

$x+5= \sqrt{x} ( \sqrt{x}+1) +1 ( \sqrt{x}+1)$

$x+5= x+ \sqrt{x}+ \sqrt{x}+1$

$x+5= x+ 2 \sqrt{x}+1$

Subtract 2√x, x, and 5 on both sides.

$x- 2 \sqrt{x} -x= 1-5$

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

Cancel negative signs.

$2\sqrt{x} = 4$

Divide both sides by 2.

$\sqrt{x} =2$

Square both sides.

$x=2^2$

$x=4$

Check if the solution in the equation works.

$\sqrt{x+5} = \sqrt{x}+1$

Let $x=4$

$\sqrt{4+5} = \sqrt{4}+1$

$\sqrt{9} = 2+1$

$3=3$

The value of x as 4 works in the equation.