extraneous equation 10 points?

sqrt(x + 42) = x

Square both sides of the equation:

x + 42 = x^2

Subtract x + 42 from both sides:

0 = x^2 - x - 42

x^2 - x - 42 = 0

Factor:

(x - 7)(x + 6) = 0

x - 7 = 0

x = 7

x + 6 = 0

x = -6

x = -6 or 7

CHECK:

sqrt(-6 + 42) = -6

sqrt(36) = -6

sqrt(36) = +/- 6

So, x = -6 works.

sqrt(7 + 42) = 7

sqrt(49) = 7

sqrt(49) = +/- 7

So, x = 7 works.

(x+42)^.5 = x

square both sides:

x + 42 = x^2

x^2 - x - 42 = 0

(x-7)(x+6) = 0

x = 7, x = -6

Check:

(7+42)^.5 = 7 = x

(-6+42)^.5 = 6 does not equal = -6.

x=7 is a solution, x=-6 is extraneous.

(sqrt(x+42))^2 = x^2

x + 42 = x^2

x + 42 - x - 42 = x^2 - x - 42

0 = x^2 - x - 42

now this is a quadratic equation

x^2 - x - 42 = 0

(x+6)(x-7) = 0

so x=-6 or x=7

now we check

sqrt(-6+42)=-6

sqrt(36)=-6

-6=-6

(-6*-6=36 so the square root of 36 could be 6 or -6)

thats good

sqrt(7+42)=7

sqrt(49)=7

7=7

thats good

so x=-6 or x=7

sqrt(x+42) = x

x^2 - x - 42 = 0

(x-7)(x+6) = 0

x = -6 or x = 7

x=-6 doesn't satisfy the original equation (we got an extra solution by squaring the equation which, strictly speaking, is NOT an equivalent operation)

so x=7 is the rolution