## help2013 Group Title solve inequality 10 months ago 10 months ago

1. help2013

im writing info

2. help2013

$x-4=\sqrt{x+8}$

3. bii17

x will be equal to 1

4. help2013

can u show me how plzzzzz! an dthanks

5. bii17

$(x-4)^2 = x+8$ $x^2 -8x +16 = x+8$ combine similar terms it will become $x^2 -9x+8 =0$ factor (x+9) (x-1) = 0 to check, substitute x=-9 or x=1

6. help2013

thanks:)

7. tkhunny

$$x-4 = \sqrt{x+8}$$ First, before ANYTHING else, think about the Domain. $$x+8\ge 0\;or\;x \ge -8$$ You tell me why? $$x-4\ge 0\;or\;x \ge 4$$ You tell me why? After that, if we get ANY result for x that is less than 4, we will simply discard it as extraneous. It should not even be considered as a solution. x = 1 is not correct. x = -9 is not correct.

8. help2013

yeah thanks:)

9. tkhunny

Seriously. Go try them in the original equation. They fail miserably.

10. help2013

well then is the answer zero

11. tkhunny

No, x = 0 is also incorrect. The only suspected results are x = 1 and x = 8. (The factoring shown above was incorrect.) x = 1 is immediately discarded as not in the Domain. x = 8 is a solution. x = -9 could not have been a solutions, as -9 < 1. x = 0 could not be a solution, as 0 < 1 If you rule out ALL your suspects, there is NO solution. In this case x = 8 is the winner.

12. help2013

ohh well thats not one of my answer choices?:/

13. help2013

opps nv thanks for the help