Question

Incase you can’t see the picture, it says “Explain how to solve the following equation and check for extraneous solutions: x= square root of -2x+6 then out of square root +3 Please help guys

Answer 1

x= 3 ; (x=1 is an extraneous solution)

Explanation:

-2x+6 is under a square root, so its value must be > or equal to 0

If we solve the inequality

-2x+6 ≥ 0 we obtain

-x+3≥ 0

-x ≥ -3

x ≤ 3

now that we know the possible values of x we can proceed like this:

1) move 3 on the left side of equal and change its sign

x-3 = √-2x+6

2) square the two members of the equation

x^2 + 9 -6x = -2x+6

3) solve the equation

x^2 -4x +3 = 0

(x-3)(x-1) = 0

x = 3

x = 1

4) Check if the solution are ≤ 3 and if they make true the equation

Both solutions are ≤ 3

if x = 3

3 = √-6+6 + 3

3 = 3

true

x=1

1 = √-2+6 + 3

1 = 2+3

1 = 5 false

In conclusion we can say the 3 is a real solution, while 1 is an extraneous solution

Answer 2

x=3

Step-by-step explanation: too much math to put here