Question

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

Answer 1

Answer:

$x=1$

Step-by-step explanation:

$\sqrt{x-1} =2-\sqrt{x+3}$

Square both sides.

$\left(\sqrt{x-1}\right)^2=\left(2-\sqrt{x+3}\right)^2$

$x-1=x+7-4\sqrt{x+3}$

Subtract x on both sides.

$x-1-x=x+7-4\sqrt{x+3}-x$

$-1=-4\sqrt{x+3}+7$

Subtract 7 on both sides.

$-1-7=-4\sqrt{x+3}+7-7$

$-8=-4\sqrt{x+3}$

Square both sides.

$\left(-8\right)^2=\left(-4\sqrt{x+3}\right)^2$

$64=16x+48$

Subtract 48 on both sides.

$64-48=16x+48-48$

$16=16x$

Divide both sides by 16.

$\frac{16}{16} =\frac{16x}{16}$

$1=x$

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