Question

Solving square root and other radical equations..
problem is below

Answer 1

\[\sqrt{3x+3} -1=x\]

Answer 2

\[\sqrt{x+6} +2= x+6\]

Answer 3

@jim_thompson5910

Answer 4

@Mertsj

Answer 5

@Hero

Answer 6

\[\large \sqrt{3x+3} -1=x\]


\[\large \sqrt{3x+3}=x+1\]


\[\large (\sqrt{3x+3})^2=(x+1)^2\]


\[\large 3x+3=(x+1)^2\]


\[\large 3x+3=x^2 + 2x+1\]


\[\large 0=x^2 + 2x+1-3x-3\]


\[\large x^2 - x - 2 = 0\]


I'll let you finish

Answer 7

ok what should i do after, isnt that the final answer, and i wow u make it so much easier

Answer 8

no that's not the last step, but it's getting there

Answer 9

you need to solve for x

then you need to check the answer(s)

Answer 10

oh so in this case would you factor?

Answer 11

that's one way to do it

Answer 12

so i got x=2 and x+-1

Answer 13

x=-1

Answer 14

good, x = 2 or x = -1

Answer 15

check both solutions now

Answer 16

ok thank you and for the other problem it seems a lil bit more confusing

Answer 17

same idea just different numbers

Answer 18

\[\large \sqrt{x+6} +2= x+6\]


\[\large \sqrt{x+6}= x+6-2\]


\[\large \sqrt{x+6}= x+4\]


\[\large (\sqrt{x+6})^2 = (x+4)^2\]


\[\large x+6 = (x+4)^2\]


\[\large x+6 = x^2 + 8x + 16\]


I'll stop here, but there's more to do.

Answer 19

so the last part so far you would subtract x-6 on both sides and then factor

Answer 20

yes, that gives you


\[\large x+6 = x^2 + 8x + 16\]


\[\large 0 = x^2 + 8x + 16 - x - 6\]


\[\large x^2 + 7x + 10 = 0\]

Answer 21

so i guess for this one i have to use the quadratic formula, right?

Answer 22

no you can factor

Answer 23

\[\large x^2 + 7x + 10 = 0\]

\[\large (x+5)(x+2) = 0\]

Answer 24

ohhhh wow i was looking at it a different way lol, ok well thanks i totally understand it now, thanks so much for your help :)