When solved correctly, the equation 4x−8−−−−−√=x−5 has solutions of x=3 and x=11.

Which, if any, of the solutions are extraneous?

x = 3 is extraneous

Step-by-step explanation:

Given

= x - 5 ( square both sides to clear the radical )

4x - 8 = (x - 5)² ← expand using FOIL

4x - 8 = x² - 10x + 25 ( subtract 4x - 8 from both sides )

0 = x² - 14x + 33 ← in standard form

0 = (x - 11)(x - 3) ← in factored form

Equate each factor to zero and solve for x

x - 11 = 0 ⇒ x = 11

x - 3 = 0 ⇒ x = 3

As a check

Substitute these values into the equation and if both sides are equal then they are solutions

x = 11

left side = = = 6

right side = 11 - 5 = 6

Thus x = 11 is a solution

x = 3

left side = = = 2

right side = 3 - 5 = - 2 ≠ 2

Thus x = 3 is an extraneous solution

but which class are u may know that first

2,160 π

step-by-step explanation: