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:
[tex]\boxed{x=4}[/tex]
Step-by-step explanation:
[tex]\sqrt{x+5} = \sqrt{x}+1[/tex]
Take the square on both sides.
[tex]x+5=( \sqrt{x}+1)^2[/tex]
Expand brackets.
[tex]x+5=( \sqrt{x}+1) ( \sqrt{x}+1)[/tex]
[tex]x+5= \sqrt{x} ( \sqrt{x}+1) +1 ( \sqrt{x}+1)[/tex]
[tex]x+5= x+ \sqrt{x}+ \sqrt{x}+1[/tex]
[tex]x+5= x+ 2 \sqrt{x}+1[/tex]
Subtract 2√x, x, and 5 on both sides.
[tex]x- 2 \sqrt{x} -x= 1-5[/tex]
[tex]-2 \sqrt{x} = -4[/tex]
Cancel negative signs.
[tex]2\sqrt{x} = 4[/tex]
Divide both sides by 2.
[tex]\sqrt{x} =2[/tex]
Square both sides.
[tex]x=2^2[/tex]
[tex]x=4[/tex]
Check if the solution in the equation works.
[tex]\sqrt{x+5} = \sqrt{x}+1[/tex]
Let [tex]x=4[/tex]
[tex]\sqrt{4+5} = \sqrt{4}+1[/tex]
[tex]\sqrt{9} = 2+1[/tex]
[tex]3=3[/tex]
The value of x as 4 works in the equation.
