Misaka solved the radical equation x-3=√4x-7 but did not check her solutions. which shows the true solution(s) to the radical equation? a. x=2 b. x=8 c. x=2, and x=8 d. there are no true solutions to the equation

B

Step-by-step explanation:

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

i dont have an answer

step-by-step explanation:

need more information

step-by-step explanation: