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.

x = 3 is a true solution.

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

Step-by-step 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:

√(3-2)-4=3-6

√1 -4 = -3

 1 - 4 = -3

    -3 = -3

√(2-2)-4=2-6

0 - 4 = -4

    -4 = -4

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2