## dddan how do sove for the solutions for ; the square root of x+ the square root of x-36 which then =0? one year ago one year ago

1. tejasree

Subtract the square root of x-36 from both sides! now: the square root of x = -the square root of x-36

2. tejasree

$\sqrt{x} = -\sqrt{x-36}$

3. dddan

what about the 2?

4. tejasree

Square both of them$(\sqrt{x})^{2} = (-\sqrt{x-36})^{2}$

5. tejasree

what 2

6. tejasree

?

7. dddan

im so sry the whole thing is equal to 2 not 0 im so sry

8. tejasree

ok...then it is: $(\sqrt{x})^{2} = (2-\sqrt{x-36})^{2}$

9. tejasree

$x = 4 - 2\sqrt{x-36} - 2\sqrt{x-36} + x - 36$

10. tejasree

$x = -32 + x + -4\sqrt{x-36}$

11. tejasree

$x - x + 32 = -4\sqrt{x-36}$

12. tejasree

$32 = -4\sqrt{x-36}$

13. tejasree

$(32)^{2} = (-4\sqrt{x-36})^{2}$

14. tejasree

no no!!

15. tejasree

wait a sec!

16. tejasree

Instead of square rooting 32 and the other, first divide -4 from both sides!

17. tejasree

$(-8)^{2} = (\sqrt{x-36})^{2}$

18. tejasree

you following?

19. tejasree

64 = X-36 64 + 36 = X - 36 + 36 100 = X

20. tejasree

thank you! :)