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anonymous
 one year ago
I need help with algebra again
anonymous
 one year ago
I need help with algebra again

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{6x+1}5=7\sqrt{3x+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1ok get the numbers to the right and those square roots to the left

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: we can rewrite your equation as below: \[\Large 7  5 = \sqrt {6x + 1}  \sqrt {3x + 1} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1half of me is like square both sides to get rid of those square roots. after that step

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2after a first square, we get: \[\Large 4 = 6x + 1 + 3x + 1  2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \] please simplify @MathHater82

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0ok but how did you do the step before that

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large 2= \sqrt {6x + 1}  \sqrt {3x + 1} \] \[\Large 2^2= (\sqrt {6x + 1}  \sqrt {3x + 1})^2 \] \[\Large 4= (\sqrt {6x + 1}  \sqrt {3x + 1})(\sqrt {6x + 1}  \sqrt {3x + 1}) \] and then expand ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since I have computed the square of both sides, like this: \[\Large {\left( {7  5} \right)^2} = {\left( {\sqrt {6x + 1}  \sqrt {3x + 1} } \right)^2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1good luck to that on the right hand side that's killer

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next, we can write: \[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x  2\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and, 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
 one year ago
Best ResponseYou've already chosen the best response.2or, after a simplification: \[\Large 4\left( {18{x^2} + 9x + 1} \right) = 81{x^2} + 4  36x\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I have applied the foil method inside the radical at left side

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, please continue @MathHater82

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1oh yes it's getting to be a bit easier. . . . after those square roots are gone.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok im just trying to follow this all on my white board i have here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[36x+4=9x^236x+4\] \[72x=9x^2\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! correct! now try to solve that equation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: 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
 one year ago
Best ResponseYou've already chosen the best response.1maybe we can shift the 9x^2 and se the equation equal to 0. whoops backwards 0 = 9x^272x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yes solve for x after what @Michele_Laino typed there's 2 x's to solve.. so first we have 9x=0 and then x8=0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2x=8 is correct! nevertheless also x=0 is a solution that's right! @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1x = 0,8 but uh oh only one x works

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, x=0 is a solution or not? @MathHater82

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please recall this step: \[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x  2\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I think plugging in x = 0 in the equation that we had started with is easier.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2noew, 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
 one year ago
Best ResponseYou've already chosen the best response.2or: \[\Large x > \frac{9}{2}\] now x=0, doesn't check that condition, so x=0 can not be an acceptable solution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437567359392:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1also when x = 0 \[\sqrt{1}5=7\sqrt{1}\] 15=71 \[6 \neq 8\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! that is another possible way to solve the question

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2mathematicians call x=0 an extraneous solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't really understand why x=0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since x=0 and x=8 are the solutions of this equation: \[\Large 9{x^2}  72x = 0\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1we obtained two solutions, but 1 works.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did we obtain the solution 0 and which one worked

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1the equation \[\Large 9{x^2}  72x = 0 \] got factored.. we can factor a 9x out

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\large 9x(x8)=0\] then we solve for each case 9x=0 and x8=0 that leads us to x =0,8 but only 1 works.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2furthermore. 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
 one year ago
Best ResponseYou've already chosen the best response.0ok cool i get it thanks for the help
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