Question

Which statements describe the solutions to (√x-2)-4=x-6? Check all that apply.

There are no true solutions to the radical equation.
x = 2 is an extraneous solution.
x = 3 is a true solution.
There is only 1 true solution to the equation.
The zeros of 0 = x2 – 5x + 6 are possible solutions to the radical equation.

Answer 1

C. x = 3 is a true solution.

E. The zeros of 0 = x2 – 5x + 6 are possible solutions to the radical equation.

Explanation:

Answer 2

x = 3 is a true solution.

The zeros of 0 = x2 – 5x + 6 are possible solutions to the radical equation.

Explanation:

Given in the question an equation,

√(x-2) - 4 = x - 6

√(x-2) = x - 6 + 4

√(x-2) = x -2

Take square on both sides of the equation

√(x-2)² = (x -2)²

x - 2 = x² - 4x + 4

0 = x² - 4x - x + 4 + 2

0 = x² - 5x + 6

a = 1

b = -5

c = 6

x = -b±√(b²-4ac) / 2a

x = -(-5)±√(5²-4(1)(6)) / 2(1)

x = 5 ± √1 / 2

x = 5 + 1 / 2 or 5 - 1 /2

x = 3 or x = 2

Plug value of x in the radical equation:

x=3

√(3-2)-4=3-6

√1 -4 = -3

1 - 4 = -3

-3 = -3

x=2

√(2-2)-4=2-6

0 - 4 = -4

-4 = -4