√(79-5x) = 6x+4
79-5x = (6x+4)^2 = 36x^2+48x+16
36x^2+53x-63 = 0
(4x+9)(9x-7) = 0
x = -9/4 or 7/9
but, you have to check for spurious roots:
√(79-5(-9/4)) = √(79+45/4) = 9.5
but, 6(-9/4)+4 = -9.5
This is the spurious root. If you square both sides, it becomes true, but not in the original.
√(79-5(7/9)) = 25/3
6(7/9)+4 = 25/3
So, 7/9 is the only solution
Solve
square root of 79-5x =6x+4
(Keep your answers in simplified improper fraction form)
1 answer