Solving $8x-3+sqrt{x+2}-sqrt{x-1}=7 sqrt{x^2+x-2}$












3












$begingroup$



Solve the equation
$$8x-3+sqrt{x+2}-sqrt{x-1}=7 sqrt{x^2+x-2}$$




I have this idea: set $$sqrt{x+2}=a , x+2=a^2 , sqrt{x-1}=b.$$
So $$x-1=b^2 , 2a^2+6b^2 =8b-4$$ and $$x^2+x-2 =a^2b^2$$ and then I'd simplify, but it's still very hard to solve.
Any hint is appreciated.






algebra-precalculus problem-solving radicals substitution algebraic-equations







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  • 2




    $begingroup$
    Please use Mathjax!
    $endgroup$
    – Le Anh Dung
    Jan 14 at 3:26






  • 3




    $begingroup$
    Your substitutions seem correct. One last thing though: Can $8x-3$ be written as a linear combination of $x+2$ and $x-1$? I.e: can we write $8x-3$ as $k(x+2) + l(x-1)$, where $k$ and $l$ are integers?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 14 at 3:35
















3












$begingroup$



Solve the equation
$$8x-3+sqrt{x+2}-sqrt{x-1}=7 sqrt{x^2+x-2}$$




I have this idea: set $$sqrt{x+2}=a , x+2=a^2 , sqrt{x-1}=b.$$
So $$x-1=b^2 , 2a^2+6b^2 =8b-4$$ and $$x^2+x-2 =a^2b^2$$ and then I'd simplify, but it's still very hard to solve.
Any hint is appreciated.






algebra-precalculus problem-solving radicals substitution algebraic-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Please use Mathjax!
    $endgroup$
    – Le Anh Dung
    Jan 14 at 3:26






  • 3




    $begingroup$
    Your substitutions seem correct. One last thing though: Can $8x-3$ be written as a linear combination of $x+2$ and $x-1$? I.e: can we write $8x-3$ as $k(x+2) + l(x-1)$, where $k$ and $l$ are integers?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 14 at 3:35














3












3








3


2



$begingroup$



Solve the equation
$$8x-3+sqrt{x+2}-sqrt{x-1}=7 sqrt{x^2+x-2}$$




I have this idea: set $$sqrt{x+2}=a , x+2=a^2 , sqrt{x-1}=b.$$
So $$x-1=b^2 , 2a^2+6b^2 =8b-4$$ and $$x^2+x-2 =a^2b^2$$ and then I'd simplify, but it's still very hard to solve.
Any hint is appreciated.






algebra-precalculus problem-solving radicals substitution algebraic-equations







share|cite|improve this question











$endgroup$





Solve the equation
$$8x-3+sqrt{x+2}-sqrt{x-1}=7 sqrt{x^2+x-2}$$




I have this idea: set $$sqrt{x+2}=a , x+2=a^2 , sqrt{x-1}=b.$$
So $$x-1=b^2 , 2a^2+6b^2 =8b-4$$ and $$x^2+x-2 =a^2b^2$$ and then I'd simplify, but it's still very hard to solve.
Any hint is appreciated.






algebra-precalculus problem-solving radicals substitution algebraic-equations




algebra-precalculus problem-solving radicals substitution algebraic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 5:21









Michael Rozenberg

101k1591193




101k1591193










asked Jan 14 at 3:24









HeartHeart

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27017








  • 2




    $begingroup$
    Please use Mathjax!
    $endgroup$
    – Le Anh Dung
    Jan 14 at 3:26






  • 3




    $begingroup$
    Your substitutions seem correct. One last thing though: Can $8x-3$ be written as a linear combination of $x+2$ and $x-1$? I.e: can we write $8x-3$ as $k(x+2) + l(x-1)$, where $k$ and $l$ are integers?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 14 at 3:35














  • 2




    $begingroup$
    Please use Mathjax!
    $endgroup$
    – Le Anh Dung
    Jan 14 at 3:26






  • 3




    $begingroup$
    Your substitutions seem correct. One last thing though: Can $8x-3$ be written as a linear combination of $x+2$ and $x-1$? I.e: can we write $8x-3$ as $k(x+2) + l(x-1)$, where $k$ and $l$ are integers?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 14 at 3:35








2




2




$begingroup$
Please use Mathjax!
$endgroup$
– Le Anh Dung
Jan 14 at 3:26




$begingroup$
Please use Mathjax!
$endgroup$
– Le Anh Dung
Jan 14 at 3:26




3




3




$begingroup$
Your substitutions seem correct. One last thing though: Can $8x-3$ be written as a linear combination of $x+2$ and $x-1$? I.e: can we write $8x-3$ as $k(x+2) + l(x-1)$, where $k$ and $l$ are integers?
$endgroup$
– Aniruddh Venkatesan
Jan 14 at 3:35




$begingroup$
Your substitutions seem correct. One last thing though: Can $8x-3$ be written as a linear combination of $x+2$ and $x-1$? I.e: can we write $8x-3$ as $k(x+2) + l(x-1)$, where $k$ and $l$ are integers?
$endgroup$
– Aniruddh Venkatesan
Jan 14 at 3:35










1 Answer
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$begingroup$

The domain gives $xgeq1$ and we need to solve that
$$sqrt{x+2}-sqrt{x-1}=7sqrt{x^2+x-2}-8x+3.$$
Now, since $sqrt{x+2}-sqrt{x-1}geq0,$ we obtain
$$7sqrt{x^2+x-2}-8x+3geq0$$ or
$$frac{97-sqrt{2989}}{30}leq xleqfrac{97+sqrt{2989}}{30}.$$
Thus, we need to solve
$$left(sqrt{x+2}-sqrt{x-1}right)^2=left(7sqrt{x^2+x-2}-(8x-3)right)^2$$ or
$$(112x-44)sqrt{x^2+x-2}=113x^2-x-90$$ or
$$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or
$$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or
$$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer:
$$left{2,frac{23+4sqrt{10}}{9}right}.$$
Also, your idea helps.



Indeed, we got the following system:
$$a^2-b^2=3$$ and
$$8(a^2-2)-3+a-b=7ab.$$
From the second equation we obtain:
$$b=frac{8a^2+a-19}{7a+1},$$ which gives
$$a^2-left(frac{8a^2+a-19}{7a+1}right)^2=3$$ or
$$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.






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    3












    $begingroup$

    The domain gives $xgeq1$ and we need to solve that
    $$sqrt{x+2}-sqrt{x-1}=7sqrt{x^2+x-2}-8x+3.$$
    Now, since $sqrt{x+2}-sqrt{x-1}geq0,$ we obtain
    $$7sqrt{x^2+x-2}-8x+3geq0$$ or
    $$frac{97-sqrt{2989}}{30}leq xleqfrac{97+sqrt{2989}}{30}.$$
    Thus, we need to solve
    $$left(sqrt{x+2}-sqrt{x-1}right)^2=left(7sqrt{x^2+x-2}-(8x-3)right)^2$$ or
    $$(112x-44)sqrt{x^2+x-2}=113x^2-x-90$$ or
    $$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or
    $$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or
    $$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer:
    $$left{2,frac{23+4sqrt{10}}{9}right}.$$
    Also, your idea helps.



    Indeed, we got the following system:
    $$a^2-b^2=3$$ and
    $$8(a^2-2)-3+a-b=7ab.$$
    From the second equation we obtain:
    $$b=frac{8a^2+a-19}{7a+1},$$ which gives
    $$a^2-left(frac{8a^2+a-19}{7a+1}right)^2=3$$ or
    $$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      The domain gives $xgeq1$ and we need to solve that
      $$sqrt{x+2}-sqrt{x-1}=7sqrt{x^2+x-2}-8x+3.$$
      Now, since $sqrt{x+2}-sqrt{x-1}geq0,$ we obtain
      $$7sqrt{x^2+x-2}-8x+3geq0$$ or
      $$frac{97-sqrt{2989}}{30}leq xleqfrac{97+sqrt{2989}}{30}.$$
      Thus, we need to solve
      $$left(sqrt{x+2}-sqrt{x-1}right)^2=left(7sqrt{x^2+x-2}-(8x-3)right)^2$$ or
      $$(112x-44)sqrt{x^2+x-2}=113x^2-x-90$$ or
      $$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or
      $$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or
      $$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer:
      $$left{2,frac{23+4sqrt{10}}{9}right}.$$
      Also, your idea helps.



      Indeed, we got the following system:
      $$a^2-b^2=3$$ and
      $$8(a^2-2)-3+a-b=7ab.$$
      From the second equation we obtain:
      $$b=frac{8a^2+a-19}{7a+1},$$ which gives
      $$a^2-left(frac{8a^2+a-19}{7a+1}right)^2=3$$ or
      $$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        The domain gives $xgeq1$ and we need to solve that
        $$sqrt{x+2}-sqrt{x-1}=7sqrt{x^2+x-2}-8x+3.$$
        Now, since $sqrt{x+2}-sqrt{x-1}geq0,$ we obtain
        $$7sqrt{x^2+x-2}-8x+3geq0$$ or
        $$frac{97-sqrt{2989}}{30}leq xleqfrac{97+sqrt{2989}}{30}.$$
        Thus, we need to solve
        $$left(sqrt{x+2}-sqrt{x-1}right)^2=left(7sqrt{x^2+x-2}-(8x-3)right)^2$$ or
        $$(112x-44)sqrt{x^2+x-2}=113x^2-x-90$$ or
        $$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or
        $$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or
        $$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer:
        $$left{2,frac{23+4sqrt{10}}{9}right}.$$
        Also, your idea helps.



        Indeed, we got the following system:
        $$a^2-b^2=3$$ and
        $$8(a^2-2)-3+a-b=7ab.$$
        From the second equation we obtain:
        $$b=frac{8a^2+a-19}{7a+1},$$ which gives
        $$a^2-left(frac{8a^2+a-19}{7a+1}right)^2=3$$ or
        $$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.






        share|cite|improve this answer











        $endgroup$



        The domain gives $xgeq1$ and we need to solve that
        $$sqrt{x+2}-sqrt{x-1}=7sqrt{x^2+x-2}-8x+3.$$
        Now, since $sqrt{x+2}-sqrt{x-1}geq0,$ we obtain
        $$7sqrt{x^2+x-2}-8x+3geq0$$ or
        $$frac{97-sqrt{2989}}{30}leq xleqfrac{97+sqrt{2989}}{30}.$$
        Thus, we need to solve
        $$left(sqrt{x+2}-sqrt{x-1}right)^2=left(7sqrt{x^2+x-2}-(8x-3)right)^2$$ or
        $$(112x-44)sqrt{x^2+x-2}=113x^2-x-90$$ or
        $$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or
        $$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or
        $$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer:
        $$left{2,frac{23+4sqrt{10}}{9}right}.$$
        Also, your idea helps.



        Indeed, we got the following system:
        $$a^2-b^2=3$$ and
        $$8(a^2-2)-3+a-b=7ab.$$
        From the second equation we obtain:
        $$b=frac{8a^2+a-19}{7a+1},$$ which gives
        $$a^2-left(frac{8a^2+a-19}{7a+1}right)^2=3$$ or
        $$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 14 at 5:20

























        answered Jan 14 at 5:06









        Michael RozenbergMichael Rozenberg

        101k1591193




        101k1591193






























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