Question

Could someone help me out with this..?
find the non-extraneous solutions of...

Answer 1

x = −2

x = −6

x = −6 and x = −2

x = 2

Answer 2

Do you know how to rewrite equations and find roots?

Answer 3

could you help me with that..?

Answer 4

Alright so first you want to do is get rid of the square root on the left, so you square everything \[x+6+25=x ^{2}+1\]

Then you rearrange everything to get the general form \[ax ^{2}+bx+c=0\]
so you get \[x ^{2}-x-30=0\]

Now, there are 2 ways of solving quadratic equations. Do you know them?

Answer 5

no :// so sorry

Answer 6

i gave the problem a try, I got x = x2 +12x -30
but that's probably completely wrong...well it is..

Answer 7

If not, this might help: http://www.mathwarehouse.com/quadratic/solve-quadratic-equation.php Sorry I made a mistake. There are 4 ways. I can't explain them myself because I have to go, but you should be able to solve it with the help of that page.

Now, you will get 2 values for x. Non-extraneous solutions are solutions that actually agree with the equation. That is, if you put the value in the equation, the left hand side and right hand side will agree with each other :) Good luck and I'll help you when I get back if you still need help.

Answer 8

okay thank you! so it would -6 and 2!!

Answer 9

got it!

Answer 10

Ah so your problem was when you rearranged the equation?