anonymous
  • anonymous
I need help with algebra again
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[-\sqrt{6x+1}-5=-7-\sqrt{3x+1}\]
UsukiDoll
  • UsukiDoll
thinking
anonymous
  • anonymous
lol ok

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UsukiDoll
  • UsukiDoll
ok get the numbers to the right and those square roots to the left
Michele_Laino
  • Michele_Laino
hint: we can rewrite your equation as below: \[\Large 7 - 5 = \sqrt {6x + 1} - \sqrt {3x + 1} \]
UsukiDoll
  • UsukiDoll
half of me is like square both sides to get rid of those square roots. after that step
Michele_Laino
  • 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
  • UsukiDoll
oh golly .
anonymous
  • anonymous
o.o im lost
Michele_Laino
  • 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
  • anonymous
ok but how did you do the step before that
UsukiDoll
  • 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
  • UsukiDoll
with foil?
Michele_Laino
  • 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
  • anonymous
oh ok
UsukiDoll
  • UsukiDoll
good luck to that on the right hand side that's killer
Michele_Laino
  • Michele_Laino
next, we can write: \[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]
Michele_Laino
  • 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
  • Michele_Laino
or, after a simplification: \[\Large 4\left( {18{x^2} + 9x + 1} \right) = 81{x^2} + 4 - 36x\]
Michele_Laino
  • Michele_Laino
I have applied the foil method inside the radical at left side
Michele_Laino
  • Michele_Laino
so, please continue @MathHater82
UsukiDoll
  • UsukiDoll
oh yes it's getting to be a bit easier. . . . after those square roots are gone.
anonymous
  • anonymous
ok im just trying to follow this all on my white board i have here
anonymous
  • anonymous
give me a sec
anonymous
  • anonymous
\[36x+4=9x^2-36x+4\] \[72x=9x^2\]
Michele_Laino
  • Michele_Laino
ok! correct! now try to solve that equation
anonymous
  • anonymous
x=8
Michele_Laino
  • 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
  • UsukiDoll
maybe we can shift the 9x^2 and se the equation equal to 0. whoops backwards 0 = 9x^2-72x
UsukiDoll
  • 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
  • Michele_Laino
x=8 is correct! nevertheless also x=0 is a solution that's right! @UsukiDoll
UsukiDoll
  • UsukiDoll
x = 0,8 but uh oh only one x works
Michele_Laino
  • Michele_Laino
so, x=0 is a solution or not? @MathHater82
Michele_Laino
  • Michele_Laino
please recall this step: \[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]
UsukiDoll
  • UsukiDoll
I think plugging in x = 0 in the equation that we had started with is easier.
Michele_Laino
  • 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
  • Michele_Laino
now*
Michele_Laino
  • 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
  • Michele_Laino
|dw:1437567359392:dw|
UsukiDoll
  • UsukiDoll
also when x = 0 \[-\sqrt{1}-5=-7-\sqrt{1}\] -1-5=-7-1 \[-6 \neq -8\]
Michele_Laino
  • Michele_Laino
yes! that is another possible way to solve the question
Michele_Laino
  • Michele_Laino
mathematicians call x=0 an extraneous solution
anonymous
  • anonymous
I don't really understand why x=0
Michele_Laino
  • Michele_Laino
since x=0 and x=8 are the solutions of this equation: \[\Large 9{x^2} - 72x = 0\]
UsukiDoll
  • UsukiDoll
we obtained two solutions, but 1 works.
anonymous
  • anonymous
how did we obtain the solution 0 and which one worked
UsukiDoll
  • UsukiDoll
the equation \[\Large 9{x^2} - 72x = 0 \] got factored.. we can factor a 9x out
UsukiDoll
  • 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
  • anonymous
ok
Michele_Laino
  • 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
  • anonymous
ok cool i get it thanks for the help

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