(√x+10 ) + (√x-6) = 8 pls show work

- anonymous

(√x+10 ) + (√x-6) = 8 pls show work

- chestercat

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- anonymous

Add like terms

- anonymous

2 sqrt[x]+10-6

- anonymous

2 sqrt[x]+4=8
2 sqrt[x]=4
(sqrt[x])^2=(2)^2
x=4

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## More answers

- yuki

littleice, is this the problem you need help with ?

- anonymous

i know the answer comes to 15

- anonymous

yes thanks

- yuki

let me ask you something first
is it \[\sqrt x +10\]
or
\[\sqrt{x+10}\]

- anonymous

\[\sqrt{x+10}+\sqrt{x-6}=8\]

- yuki

okay,
so our main goal is to get rid of the square roots.
but before we proceed, there are things that you want
to always remember about eqn.s that involves sqrt.
it is that there are x's that are not allowed to use.
since te number inside the radical has to be positive,
first we have to note that
x>-10 and x>6

- anonymous

i thought it was x cant equal -10 or 6?

- yuki

if we get an answer that doesn't fall in this category, then there are no solutions.
another thing that you have to do, (I'm not going to go into the detail unless you want to) is to check the answers when you get it.
the gist of the reason has to do with the fact that\[x^2=k, means, x= \pm \sqrt{k}\]

- yuki

-10 and 6 are actually ok.
because sqrt(0) = 0
but sqrt(x) when x<0, becomes and imaginary number

- anonymous

k

- yuki

alright let's proceed

- yuki

when you get rid of sqrt.s you want to square them.
but when you square both sides just as it is, you will
end up having a square root again, so to avoid that,
you will isolate one of the terms that has a square root
around it, and deal with that first.
as follows

- yuki

\[(\sqrt{x+10})^2 = (8-\sqrt{x-6})^2\]
so\[x+10 = 64 +16\sqrt{x-6}+(x-6)\]

- yuki

now you only have one square root, so you isolate it
and then square both sides again
\[({x + 10 -64 -x +6 \over 16})^2 = \sqrt{x-6}^2\]

- yuki

that way you will have a quadratic eqn. and
you know how to solve it from there.
don't forget to check the solutions, by plugging them in
because sometimes they will fail.

- yuki

\[(-3 )^2= \sqrt(x-6)^2\]
is what you will get when you simplify the above
so 9 = x -6
thus x =15

- yuki

however, let's see what happens when we plug this back in
into the original eqn.
sqrt(15+10)+sqrt(15-6) =5+3 =8
which actually works.
now we can guarantee that x =15.

- yuki

did it help at all ?

- anonymous

idk what went thru my head but i got confused on how u did it did it a weird way and got the answer aswell

- yuki

oh... ok :(
let me know if you need any explanations though :)

- anonymous

heres how i did it
\[\left( \sqrt{x+10} \right)^{2} = (8-\sqrt{x-6})^{2}\]
\[x+10=64+16\sqrt{x-6} + \sqrt{x-6}\]
\[x+10-64-x-6=16\sqrt{x-6}\]
\[48^{2}=(16\sqrt{x-6})^{2}\]
2304=256x-6
2304/256=x-6
9=x-6
15=x

- yuki

it is the same thing

- yuki

I just divided 16 on both sides first

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