I need help with algebra again

- anonymous

I need help with algebra again

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

\[-\sqrt{6x+1}-5=-7-\sqrt{3x+1}\]

- UsukiDoll

thinking

- anonymous

lol ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- UsukiDoll

ok get the numbers to the right and those square roots to the left

- Michele_Laino

hint:
we can rewrite your equation as below:
\[\Large 7 - 5 = \sqrt {6x + 1} - \sqrt {3x + 1} \]

- UsukiDoll

half of me is like square both sides to get rid of those square roots. after that step

- Michele_Laino

after a first square, we get:
\[\Large 4 = 6x + 1 + 3x + 1 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \]
please simplify @MathHater82

- UsukiDoll

oh golly .

- anonymous

o.o im lost

- Michele_Laino

hint:
here are the next steps:
\[\Large \begin{gathered}
4 = 9x + 2 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\
\hfill \\
2 = 9x - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\
\end{gathered} \]

- anonymous

ok but how did you do the step before that

- UsukiDoll

\[\Large 2= \sqrt {6x + 1} - \sqrt {3x + 1} \]
\[\Large 2^2= (\sqrt {6x + 1} - \sqrt {3x + 1})^2 \]
\[\Large 4= (\sqrt {6x + 1} - \sqrt {3x + 1})(\sqrt {6x + 1} - \sqrt {3x + 1}) \]
and then expand ?

- UsukiDoll

with foil?

- Michele_Laino

since I have computed the square of both sides, like this:
\[\Large {\left( {7 - 5} \right)^2} = {\left( {\sqrt {6x + 1} - \sqrt {3x + 1} } \right)^2}\]

- anonymous

oh ok

- UsukiDoll

good luck to that on the right hand side that's killer

- Michele_Laino

next, we can write:
\[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]

- Michele_Laino

and, taking the square of both sides again, we have:
\[\Large {\left( {2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} } \right)^2} = {\left( {9x - 2} \right)^2}\]

- Michele_Laino

or, after a simplification:
\[\Large 4\left( {18{x^2} + 9x + 1} \right) = 81{x^2} + 4 - 36x\]

- Michele_Laino

I have applied the foil method inside the radical at left side

- Michele_Laino

so, please continue @MathHater82

- UsukiDoll

oh yes it's getting to be a bit easier. . . . after those square roots are gone.

- anonymous

ok im just trying to follow this all on my white board i have here

- anonymous

give me a sec

- anonymous

\[36x+4=9x^2-36x+4\] \[72x=9x^2\]

- Michele_Laino

ok! correct!
now try to solve that equation

- anonymous

x=8

- Michele_Laino

hint:
we have the subsequent steps:
\[\Large \begin{gathered}
9{x^2} - 72x = 0 \hfill \\
\hfill \\
9x\left( {x - 8} \right) = 0 \hfill \\
\end{gathered} \]

- UsukiDoll

maybe we can shift the 9x^2 and se the equation equal to 0. whoops backwards
0 = 9x^2-72x

- UsukiDoll

yes solve for x after what @Michele_Laino typed
there's 2 x's to solve.. so first we have
9x=0
and then x-8=0

- Michele_Laino

x=8 is correct!
nevertheless also x=0 is a solution
that's right! @UsukiDoll

- UsukiDoll

x = 0,8 but uh oh only one x works

- Michele_Laino

so, x=0 is a solution or not? @MathHater82

- Michele_Laino

please recall this step:
\[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]

- UsukiDoll

I think plugging in x = 0 in the equation that we had started with is easier.

- Michele_Laino

noew, the left side is positive, since it is the positive square root, so also the right side has to be positive. Namely we have:
\[\Large 9x - 2 > 0\]

- Michele_Laino

now*

- Michele_Laino

or:
\[\Large x > \frac{9}{2}\]
now x=0, doesn't check that condition, so x=0 can not be an acceptable solution

- Michele_Laino

|dw:1437567359392:dw|

- UsukiDoll

also when x = 0
\[-\sqrt{1}-5=-7-\sqrt{1}\]
-1-5=-7-1
\[-6 \neq -8\]

- Michele_Laino

yes! that is another possible way to solve the question

- Michele_Laino

mathematicians call x=0 an extraneous solution

- anonymous

I don't really understand why x=0

- Michele_Laino

since x=0 and x=8 are the solutions of this equation:
\[\Large 9{x^2} - 72x = 0\]

- UsukiDoll

we obtained two solutions, but 1 works.

- anonymous

how did we obtain the solution 0 and which one worked

- UsukiDoll

the equation \[\Large 9{x^2} - 72x = 0 \] got factored.. we can factor a 9x out

- UsukiDoll

\[\large 9x(x-8)=0\]
then we solve for each case
9x=0 and x-8=0
that leads us to x =0,8 but only 1 works.

- anonymous

ok

- Michele_Laino

furthermore. looking at your original equation also these conditions have to be checked:
\[\Large \begin{gathered}
6x + 1 \geqslant 0 \hfill \\
3x + 1 \geqslant 0 \hfill \\
\end{gathered} \]
for existence of both radicals, which are automatically satisfied since:
\[\Large 8 > \frac{9}{2}\]

- anonymous

ok cool i get it thanks for the help

Looking for something else?

Not the answer you are looking for? Search for more explanations.