Solve the equation for x. if a solution is extraneous, be sure to identify it in your final answer. square root of the quantity x minus 2 end quantity plus 8 equals x

11 only true real solution

6 is extraneous (not a solution to the original equation)

Step-by-step explanation:

sqrt(x-2)+8=x

Get the radical by itself first (the square root thing)

Subtract 8 on both sides

sqrt(x-2)=x-8

Now square both sides

x-2=(x-8)^2

x-2=x^2-16x+64  (multiply the (x-8)^2 out aka multiply (x-8) and (x-8) use foil)

Subtract x and add 2 on both sides

  0=x^2-17x+66

Find two numbers that multiply to be 66 and add to be -17

66=6(11)  so 66=-6(-11) and -6+(-11)=-17

So the factored form of the equation is

0=(x-6)(x-11)

x=6 or x=11

Check to see if any of those are bad solutions (solutions that actually don't work-aka extraneous)

Check x=6

sqrt(6-2)+8=6

sqrt(4)+8=6

2+8=6

Not right so 6 is extraneous

Check x=11

sqrt(11-2)+8=11

sqrt(9)+8=11

3+8=11

Good

So 11 is the only true real solution