anonymous one year ago Given the equation Square root of 8x plus 1 = 5, solve for x and identify if it is an extraneous solution.

1. SolomonZelman

$$\sqrt{8x+1}=5$$ or $$\sqrt{8x}+1=5$$ ?

2. anonymous

the first one :)

3. SolomonZelman

raise both sides the the second power

4. SolomonZelman

$$\large\color{red}{ \displaystyle \left(\color{black}{~\sqrt{8x+1}}~\right)^2\color{black}{=} \left(~\color{black}{5}~\right)^2 }$$

5. anonymous

okay so im super bad at math so hopefully this wont be wrong |dw:1436459928067:dw|

6. anonymous

at the end its 25

7. SolomonZelman

When you take the square root of a number, and raise it to second power you should get the number itself. Such that ( √D )² = D

8. SolomonZelman

` Side Note: I mean normally it is equivalent to |D|, but don't worry about that, you are just cancelling radicals.

9. anonymous

ooh

10. SolomonZelman

yes, so go ahed and try again please for me.

11. SolomonZelman

what do you get after raising both sides to the second power?

12. anonymous

|dw:1436460258848:dw|

13. SolomonZelman

you have 25 on the right side, and the left side is correct

14. SolomonZelman

because 5²=25, you know that right?

15. SolomonZelman

so 8x+1=25 then solve for x....

16. anonymous

|dw:1436460559771:dw|

17. SolomonZelman

yes, x=3 is correct!

18. SolomonZelman

now, to see if this solution is extraneous (if it doesn't work in original equation) or if it is not extraneous (if it DOES work in original equaion), YOU HAVE TO plug in x=3 into your original equation.

19. anonymous

thank you for the help

20. SolomonZelman

Sure...