Question

Find the non-extraneous solutions of the square root of the quantity of x plus 3 minus 4 equals x minus 1.

Answer 1

x+3=(x+3)^2
x+3=(x+3)(x+3)
x+3= x^2+6x+9
x=x^2+6x+6
0=x^2+5x+6
0=(x+3)(x+2)
x=-2
x=-3

Answer 2

x=-2 and x=-3

Explanation:

`√(x+3) -4= x-1`

Solve the equation for x

Add 4 on both sides

`√(x+3)= x+3`

Now take square on both sides

`x+3= (x+3)^2`

`x+3= (x+3)(x+3)`

`x+3= x^2+6x+9`

`x=x^2+6x+6`

`0=x^2+5x+6`

`0=(x+3)(x+2)`

x+3=0, x=-3

x+2=0, x=-2

LEts plug in -2 for x and verify the solution

`√(x+3) -4= x-1`

`√(-2+3) -4= -2-1`

`√(1) -4=-3`

`-3=-3` True

Plug in x=-3

`√(-3+3) -4= -3-1`

`√(0) -4=-4`

`-4=-4` True

So x=-2 and x=-3 are the non extraneous solution