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
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Answer:
x= 3 ; (x=1 is an extraneous solution)
Step-by-step 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