Question

Root( x-1 ) = 2 - root (x+3)

Answer 1

`x=1`

Explanation:

`√(x-1) =2-√(x+3)`

Square both sides.

`\left(√(x-1)\right)^2=\left(2-√(x+3)\right)^2`

`x-1=x+7-4√(x+3)`

Subtract x on both sides.

`x-1-x=x+7-4√(x+3)-x`

`-1=-4√(x+3)+7`

Subtract 7 on both sides.

`-1-7=-4√(x+3)+7-7`

`-8=-4√(x+3)`

Square both sides.

`\left(-8\right)^2=\left(-4√(x+3)\right)^2`

`64=16x+48`

Subtract 48 on both sides.

`64-48=16x+48-48`

`16=16x`

Divide both sides by 16.

`(16)/(16) =(16x)/(16)`

`1=x`

Switch sides.

`x=1`

Answer 2

Answer: x = 4

Explanation: root x-1 = 2- root x+3
First let’s take out the root by square both side by 2 we get:
x-1= 4-x+3
x-1 = -x+7
x+x = 7+1
2x= 8
x= 8/2=4