I have another Algebra question?

Question

You guys are great! Solve the equation identify extraneous solutions: X= the square root of 4x +32

Answer 1

x = sq rt(4x+32)
x^2 = 4x+32
x^2-4x-32=0
(x-8)(x+4)=0
x = 8, -4
8 only is the solution since
-4 not = sqrt(-16+32)

Answer 2

I like you too. square both sides to get
x^2 = 4x+ 32
By inspection, x=8, but if you solve with as a quadratic, you have x^2-4x-32 and
x= [4 +/- sqrt(16+128)]/2. which is
indicates x= -4 and 8. The -4 root is extraneous.

Answer 3

X= sqrt( 4x +32)

square both sides
x^2 = 4x + 32

x^2 - 4x - 32

factor. Look for two numbers that multiply to -32 and add to -4
-8 and 4

(x - 8) (x+4)

x = 8 or -4

check:
8 = sqrt(4*8 + 32)
8 = sqrt(64)
8 = 8

-4 is not the answer because the domian is x /< -8

Answer 4

x can equal 8 or -4

x = (4x + 32) ^ 1/2
x^2 = 4x + 32 (you get this by squaring the whole eqn)
x^2 -4x - 32 now factor this and you get
(x-8)(x+4)
then solve forx by setting the eqn to 0
x-8 = 0
x = 8

and
x+4 = 0
x = -4

if you plug 8 and -4 into the equation (one at a time) you will see it checks out, don't be fooled into thinking -4 is not a solution because -16+32 = 16 and square root of 16 is 4 and -4

Answer 5

x = sqrt( 4x + 32)
x^2 = 4x + 32
x^2 - 4x + 4 = 36
x-2 = +/- 6
x = 8 or -4

Answer 6

x=sqrt (4x+32)
x=2(sqrt x)+ 4(sqrt 2)