Question

how do sove for the solutions for ;
the square root of x+ the square root of x-36 which then =0?

Answer 1

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

Answer 2

\[\sqrt{x} = -\sqrt{x-36}\]

Answer 3

what about the 2?

Answer 4

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

Answer 5

what 2

Answer 6

?

Answer 7

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

Answer 8

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

Answer 9

\[x = 4 - 2\sqrt{x-36} - 2\sqrt{x-36} + x - 36\]

Answer 10

\[x = -32 + x + -4\sqrt{x-36}\]

Answer 11

\[x - x + 32 = -4\sqrt{x-36}\]

Answer 12

\[32 = -4\sqrt{x-36}\]

Answer 13

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

Answer 14

no no!!

Answer 15

wait a sec!

Answer 16

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

Answer 17

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

Answer 18

you following?

Answer 19

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

Answer 20

thank you! :)