What is this answer x = sqrt(x) -6?

Question

help please

Answer 1

I'm assuming that sqrt(x) covers simply x.

Therefore,

x + 6  = sqrt(x)

Squaring,

x^2 + 12x + 36 = x

x^2 + 11x + 36 = 0

which is hard to factor, but you can use the quadratic formula for it.

Unless of course it is x = sqrt(x) + 6, in which case the second guy is correct.

Answer 2

isolate the radical,

              x-6=sqrt(x)

    squaring both sides, we have

                  x^2-12x+36=x

                  x^2-13+36=0

                    (x-9)(x-4)=0

                   x=9, 4

             since 4 is an extraneous root, the solution is x=9

                        

Answer 3

x = sq rt(x)-6

x + 6 = sq rt(x)

(x+6)^2 = x

x^2 + 12x + 36 = x

x^2 + 11x + 36 = 0

Use quadratic formula to solve from there.

Answer 4

You mean x = x^1/2  -6 ,then x+6=x^1/2, then x^2 + 12x + 36 = x

x^2+11x+36=0,x has no real solution.

Answer 5

x = sq rt(x) -6



x + 6 = sq rt (x)---->YOU SQUARE BOTH SIDES

(x + 6)^2 =  (sq rt(x))^2

x^2 + 12x + 36 = x

x^2 + 11x +36 = 0-----> use the quadratic formula

a = 1 b = 11 c=36

x = -11 +/-sq rt( 11^2 - 4*1*36)/2

x= -11 +/-sq rt(-23)/2

x = [-11 +/- i sq rt(23)]/2----> the answer is a complex number with no real roots they are imaginary

Answer 6

x = √(x)-6

x + 6 = √(x)

(x+6)^2 = x

x^2 + 12x + 36 = x

x^2 + 11x + 36 = 0

x = (-11 ± √(121 - 144))/2

x = (-11 ± 23i)/2

in complex notation:

x = (-11/2) + (23i/2)

& x = (-11/2) - (23i/2)