Question

Misaka solved the radical equation x – 3 = but did not check her solutions.

(x – 3)2 =
x2 – 6x + 9 = 4x – 7
x2 – 10x + 16 = 0
(x – 2)(x – 8) = 0
x = 2 and x = 8
Which shows the true solution(s) to the radical equation x – 3 = ?

Answer 1

x - 3 =

square both sides to get rid of square root

x^2 -6x + 9 = 4x - 7 , but remember 4x - 7 can't be less than 0 since it was square roots can't be less than 0

simplify the equation

x^2 - 10x + 16= 0

x = 8, x= 2

but x cannot equal 2, because if we plug in 2, the square root will be negative,

so x can only equal 8

so B is correct.

I'm not the best at quadratics, so pls comment if something is wrong