Solve and check each radical equation. square root y+10=y-2
sqrt(y + 10) = y - 2
First thing to do is to square both sides of the equation, then solve for x:
y + 10 = (y - 2)^2
y + 10 = y^2 - 4y + 4
0 = y^2 - 5y - 6
0 = (y - 6)(y + 1)
y = 6
y = -1
Now to check, substitute the x values back to original equation.
y = -1:
sqrt(-1 + 10) = -1 - 2
sqrt(9) = -3
3 = -3
Thus x is NOT equal to -1. This is an extraneous root.
y = 6:
sqrt(6 + 10) = 6 - 2
sqrt(16) = 4
4 = 4
Thus x is indeed equal to 6.
Hope this helps~ :3