can someone solve sqrt x-5 = x-7

√(x - 5) = x - 7
square both sides
x - 5 = (x - 7)²
x - 5 = x² - 14x + 49
x² - 15x + 54 = 0
(x - 9)(x - 6) = 0
x = 6 or 9

√(x - 5) = x - 7
x - 5 = (x - 7)²
x - 5 = x² - 14x + 49
x² - 15x = - 54
x² - 15/2x = - 54 + 225/4
(x - 15/2)² = 9/4
x - 15/2 = 3/2

1st factor:
= x - 15/2 + 3/2
= x - 6

2nd factor:
= (x² - 15x + 54)/(x - 6)
= x - 9

Values of x:
x - 6 = 0, x = 6
x - 9 = 0, x = 9

Answer: x = 6, 9

Proof (x is 6):
√(6 - 5) = 6 - 7
√1 = - 1
- 1 = - 1

Proof (x is 9):
√(9 - 5) = 9 - 7
√4 = 2
2 = 2

√(x-5) = (x-7)

First square both sides to get:

x-5 = x^2-14x+49

Set everything equal to zero:

x^2-14x+49-x+5 = 0

Combine like terms:

x^2-15x+54 = 0

Solve for x:

(x-6)(x-9) = 0

x=6 and x=9

Have a good day!

well sqrt x-5 = x-7
::> x-5= (x-7)^2
::>x-5= x^2 + 49- 14x
::> x^2 +54-15x=0
::>(x-6)(x-9)=0
::> x=6 or x=9

sq both sides
x-5 = (x-7)^2
x-5 = x^2 -14x + 49
0 = x^2 -15x + 54
0 = (x-9)(x-6)
x = 9 or x =6

Square both sides: x-5=(x-7)^2
x-5=x^2-14x+49
and from there, put everything on one side and then use the quadratic equation. You already have the answer, so I won't do it for you.

√(x - 5) = x - 7
x - 5 = (x - 7)^2
x - 5 = (x - 7)(x - 7)
x - 5 = x*x - 7*x - x*7 + 7*7
x - 5 = x^2 - 14x + 49
x^2 - 14x - x + 5 + 49 = 0
x^2 - 15x + 54 = 0
x^2 - 6x - 9x + 54 = 0
(x^2 - 6x) - (9x - 54) = 0
x(x - 6) - 9(x - 6) = 0
(x - 6)(x - 9) = 0

x - 6 = 0
x = 6

x - 9 = 0
x = 9

∴ x = 6 , 9

√ (x - 5) = x - 7
x - 5 = (x - 7)²
x - 5 = x² - 14x + 49
x² - 15x + 54 = 0
(x - 9)(x - 6) = 0
x = 6 , x = 9

You square both sides and you get

x-5 = x^2 - 14x + 49

shifting all the subjects to one side,

x^2 -15x + 54 = 0

With reference to formula

x = {-b +/- [rt (b^2-4ac)]} / 2a,

x = 6 or x = 9