Question

Solve. 4*sqrt(x)=sqrt(9x+9)
Honors Algebra 2 Help!

Answer 1

What do you get when you square both sides?

Answer 2

To clarify-the equation should be....\[4\sqrt{x}=\sqrt{9x+9}\]

Answer 3

you should just square the sqrt{x} and sqrt{9x+9}, correct? So it would be\[4x=9x+9\]. I have used my CAS Software and it says the answer is x=9/7. But, my teacher does no calculator tests and I don't know how to solve this by hand.

Answer 4

You square everything on both sides :)

\[ (4\sqrt x ) ^2 = ( \sqrt{9x+9} )^2 \]

Answer 5

16x = 9x + 9

Answer 6

\[ (16x) - 9x = (9x + 9) - 9x \]
\[7x = 9\]

Answer 7

OH! thanks. I have a couple more if you don't mind. I'll work it out and see if I did it right, cool?

Answer 8

for sure :)

Answer 9

\[\sqrt{x+3}=x-3\]

Answer 10

Square both sides,
\[(\sqrt{x+3})^2=(x-3)^2\]

Answer 11

mhmm :)

Answer 12

You get x+3=(x-3)^2. (x-3)^2=x^2-6x+9.
Next, you should write it out as x+3=x^2-6x+9.
Subtract x. (x+3)-x=(x^2-6x+9)-x. -----> 3=x^2-7x+9
Subract 3. (3)-3=(x^2-7x+9)-3 -----> 0=x^2-7x+6
Factor x^2-7x+6. ----->(x-1)(x-6)
Solutions could be x=1 & 6.
Check for extraneous solutions.-----both extraneous?

Answer 13

Nope, x=6. x=1 is extraneous.

Answer 14

nice!

Answer 15

I'm lost on this one. \[\sqrt{2x+15}-x=6\] @AllTehMaffs

Answer 16

plop that x from the left side to the right side, and do the same things as the last one ^_^

Answer 17

\[ (\sqrt{2x+15})^2 = (x+6)^2\]

Answer 18

Did that help? ^^