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

Answer:

x=-2 and x=-3  

Step-by-step explanation:

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

Solve the equation for x

Add 4 on both sides

$\sqrt{x+3}= x+3$

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

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

$\sqrt{-2+3} -4= -2-1$

$\sqrt{1} -4=-3$

$-3=-3$  True

Plug in x=-3

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

$\sqrt{0} -4=-4$

$-4=-4$  True

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