## anonymous 5 years ago square root of 2z+9-square root of z+6=o

1. anonymous

$\sqrt{2z+9 }-\sqrt{z+6}=0$

2. anonymous

$\sqrt{2z+9}=\sqrt{z+6}$

3. anonymous

$\implies \sqrt{2z+9}=\sqrt{z+6} \implies 2z+9=z+6 \implies z=-3$ Plug in the equation and check!!

4. anonymous

hows the work?

5. anonymous

Just substitute in the original equation with z=-3 to check it's the right solution!!

6. anonymous

7. anonymous

yea i checked!

8. anonymous

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

9. anonymous

What do you think?

10. anonymous

x-8=x-16

11. anonymous

-8 -8

12. anonymous

=8

13. anonymous

Well. The first thing you should here is to get rid of the square root. To do so, you should square both sides. Can you do that?

14. anonymous

yess

15. anonymous

you should do*

16. anonymous

Ok. Do it and show me what you get.

17. anonymous

$\sqrt({x-8})^{2}=(x-4)^{2}$

18. anonymous

$x+8=x+16$

19. anonymous

are you sure about the right hand side?

20. anonymous

(x-4)^2 is what?

21. anonymous

yes because 4x4=16

22. anonymous

all (x-4) is raised to the power of 2. Not only the 4.. So it's like (x-4) times (x-4)

23. anonymous

??

24. anonymous

:)

25. anonymous

26. anonymous

First start by squaring both sides: $(\sqrt{x+8})^2=(x-4)^2$ you did the left hand side, it's the same as x+8.. the right hand side is: $(x-4)^2=x^2-8x+16$

27. anonymous

Now, the equation will look like: $x+8=x^2-8x+16 \implies x^2-9x+8=0$ Do you know how to solve a quadratic equation?

28. anonymous

Just factorize the expression x^2-9x+8 to get: $x^2-9x+8=0 \implies (x-8)(x-1)=0 \implies x=1, x=8$ We have two values for x. Check for each one!

29. anonymous

You will find that only x=8 is a solution of the equation.