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Complicated equation, make x the subject


Making $y$ the subject of an equationTransposing an equation: x = F/k + sqrt(F/c) to get F as the subjectMake x the subject given the formula for yHow can I rearrange ($a$ * $b$)^$c$=$d$ to make $c$ the subjectRearrange the equation to make $K$ the subject: $Q=(aK^-2+bL^-2)^-1/2$Make r the Subject of the formulaRearranging angular velocity equation to make $T$ the subjectMake y the subject of $x=y^2-y$Making x the subjectFormula, making the subject.













3












$begingroup$


$$x-sqrt{x}=yx+1$$



I was ask to make x the subject, but find it very difficult.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: Isolate the square root on one side, square both sides, move x terms to one side, isolate x. You should get $$left.x= frac{-2 y-sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)},x= frac{-2 y+sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)}right.$$ Maybe a better approach is to let $t = sqrt{x}$ and you end up with a quadratic where you solve for $t$ and then substitute in $sqrt{x}$ at the end and square both sides.
    $endgroup$
    – Moo
    8 hours ago












  • $begingroup$
    I edited your title to change the word "Complicate" to "Complicated", because I assume your goal is to simplify a complicated situation rather that complicate it even more! Feel free to change it back if I'm wrong. Cheers!
    $endgroup$
    – Robert Lewis
    5 hours ago


















3












$begingroup$


$$x-sqrt{x}=yx+1$$



I was ask to make x the subject, but find it very difficult.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: Isolate the square root on one side, square both sides, move x terms to one side, isolate x. You should get $$left.x= frac{-2 y-sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)},x= frac{-2 y+sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)}right.$$ Maybe a better approach is to let $t = sqrt{x}$ and you end up with a quadratic where you solve for $t$ and then substitute in $sqrt{x}$ at the end and square both sides.
    $endgroup$
    – Moo
    8 hours ago












  • $begingroup$
    I edited your title to change the word "Complicate" to "Complicated", because I assume your goal is to simplify a complicated situation rather that complicate it even more! Feel free to change it back if I'm wrong. Cheers!
    $endgroup$
    – Robert Lewis
    5 hours ago
















3












3








3


1



$begingroup$


$$x-sqrt{x}=yx+1$$



I was ask to make x the subject, but find it very difficult.






algebra-precalculus







share|cite|improve this question











$endgroup$




$$x-sqrt{x}=yx+1$$



I was ask to make x the subject, but find it very difficult.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Robert Lewis

50.3k23369




50.3k23369










asked 8 hours ago









TriangleTriangle

854




854








  • 3




    $begingroup$
    Hint: Isolate the square root on one side, square both sides, move x terms to one side, isolate x. You should get $$left.x= frac{-2 y-sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)},x= frac{-2 y+sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)}right.$$ Maybe a better approach is to let $t = sqrt{x}$ and you end up with a quadratic where you solve for $t$ and then substitute in $sqrt{x}$ at the end and square both sides.
    $endgroup$
    – Moo
    8 hours ago












  • $begingroup$
    I edited your title to change the word "Complicate" to "Complicated", because I assume your goal is to simplify a complicated situation rather that complicate it even more! Feel free to change it back if I'm wrong. Cheers!
    $endgroup$
    – Robert Lewis
    5 hours ago
















  • 3




    $begingroup$
    Hint: Isolate the square root on one side, square both sides, move x terms to one side, isolate x. You should get $$left.x= frac{-2 y-sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)},x= frac{-2 y+sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)}right.$$ Maybe a better approach is to let $t = sqrt{x}$ and you end up with a quadratic where you solve for $t$ and then substitute in $sqrt{x}$ at the end and square both sides.
    $endgroup$
    – Moo
    8 hours ago












  • $begingroup$
    I edited your title to change the word "Complicate" to "Complicated", because I assume your goal is to simplify a complicated situation rather that complicate it even more! Feel free to change it back if I'm wrong. Cheers!
    $endgroup$
    – Robert Lewis
    5 hours ago










3




3




$begingroup$
Hint: Isolate the square root on one side, square both sides, move x terms to one side, isolate x. You should get $$left.x= frac{-2 y-sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)},x= frac{-2 y+sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)}right.$$ Maybe a better approach is to let $t = sqrt{x}$ and you end up with a quadratic where you solve for $t$ and then substitute in $sqrt{x}$ at the end and square both sides.
$endgroup$
– Moo
8 hours ago






$begingroup$
Hint: Isolate the square root on one side, square both sides, move x terms to one side, isolate x. You should get $$left.x= frac{-2 y-sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)},x= frac{-2 y+sqrt{5-4 y}+3}{2 left(y^2-2 y+1right)}right.$$ Maybe a better approach is to let $t = sqrt{x}$ and you end up with a quadratic where you solve for $t$ and then substitute in $sqrt{x}$ at the end and square both sides.
$endgroup$
– Moo
8 hours ago














$begingroup$
I edited your title to change the word "Complicate" to "Complicated", because I assume your goal is to simplify a complicated situation rather that complicate it even more! Feel free to change it back if I'm wrong. Cheers!
$endgroup$
– Robert Lewis
5 hours ago






$begingroup$
I edited your title to change the word "Complicate" to "Complicated", because I assume your goal is to simplify a complicated situation rather that complicate it even more! Feel free to change it back if I'm wrong. Cheers!
$endgroup$
– Robert Lewis
5 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

Introduce a new variable $ z= sqrt x $. Then, you get $ z^2 - z = y z^2 + 1 $ or equivalently $ (1-y) z^2 - z + 1 = 0 $. Find the solutions for $ z $. Finally, you can obtain $ x $.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Or, one might try this:



    from



    $x - sqrt x = yx + 1, tag 1$



    we may write



    $sqrt x= x - yx - 1; tag 2$



    we square:



    $ x = x^2 + y^2x^2 + 1 - 2yx^2 - 2x + 2yx$
    $= (y^2 - 2y + 1)x^2 + (2y - 2)x + 1;tag 3$



    subtract $x$:



    $(y^2 - 2y + 1)x^2 + (2y - 3)x + 1 = 0,tag 3$



    or



    $(y - 1)^2 x^2 + (2y - 3)x + 1 = 0; tag 4$



    now apply the quadratic formula:



    $x = dfrac{-(2y - 3) pm sqrt{(2y - 3)^2 - 4(y - 1)^2}}{2(y - 1)^2}; tag 5$



    simplify the radical:



    $(2y - 3)^2 - 4(y - 1)^2 = 4y^2 - 12y + 9 - 4y^2 + 8y - 4 =5 - 4y; tag 6$



    thus,



    $x = dfrac{- 2y + 3 pm sqrt{5 - 4y}}{2(y - 1)^2}. tag 7$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






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      2 Answers
      2






      active

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      active

      oldest

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      active

      oldest

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      4












      $begingroup$

      Introduce a new variable $ z= sqrt x $. Then, you get $ z^2 - z = y z^2 + 1 $ or equivalently $ (1-y) z^2 - z + 1 = 0 $. Find the solutions for $ z $. Finally, you can obtain $ x $.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Introduce a new variable $ z= sqrt x $. Then, you get $ z^2 - z = y z^2 + 1 $ or equivalently $ (1-y) z^2 - z + 1 = 0 $. Find the solutions for $ z $. Finally, you can obtain $ x $.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Introduce a new variable $ z= sqrt x $. Then, you get $ z^2 - z = y z^2 + 1 $ or equivalently $ (1-y) z^2 - z + 1 = 0 $. Find the solutions for $ z $. Finally, you can obtain $ x $.






          share|cite|improve this answer









          $endgroup$



          Introduce a new variable $ z= sqrt x $. Then, you get $ z^2 - z = y z^2 + 1 $ or equivalently $ (1-y) z^2 - z + 1 = 0 $. Find the solutions for $ z $. Finally, you can obtain $ x $.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          DunkelDunkel

          325111




          325111























              3












              $begingroup$

              Or, one might try this:



              from



              $x - sqrt x = yx + 1, tag 1$



              we may write



              $sqrt x= x - yx - 1; tag 2$



              we square:



              $ x = x^2 + y^2x^2 + 1 - 2yx^2 - 2x + 2yx$
              $= (y^2 - 2y + 1)x^2 + (2y - 2)x + 1;tag 3$



              subtract $x$:



              $(y^2 - 2y + 1)x^2 + (2y - 3)x + 1 = 0,tag 3$



              or



              $(y - 1)^2 x^2 + (2y - 3)x + 1 = 0; tag 4$



              now apply the quadratic formula:



              $x = dfrac{-(2y - 3) pm sqrt{(2y - 3)^2 - 4(y - 1)^2}}{2(y - 1)^2}; tag 5$



              simplify the radical:



              $(2y - 3)^2 - 4(y - 1)^2 = 4y^2 - 12y + 9 - 4y^2 + 8y - 4 =5 - 4y; tag 6$



              thus,



              $x = dfrac{- 2y + 3 pm sqrt{5 - 4y}}{2(y - 1)^2}. tag 7$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Or, one might try this:



                from



                $x - sqrt x = yx + 1, tag 1$



                we may write



                $sqrt x= x - yx - 1; tag 2$



                we square:



                $ x = x^2 + y^2x^2 + 1 - 2yx^2 - 2x + 2yx$
                $= (y^2 - 2y + 1)x^2 + (2y - 2)x + 1;tag 3$



                subtract $x$:



                $(y^2 - 2y + 1)x^2 + (2y - 3)x + 1 = 0,tag 3$



                or



                $(y - 1)^2 x^2 + (2y - 3)x + 1 = 0; tag 4$



                now apply the quadratic formula:



                $x = dfrac{-(2y - 3) pm sqrt{(2y - 3)^2 - 4(y - 1)^2}}{2(y - 1)^2}; tag 5$



                simplify the radical:



                $(2y - 3)^2 - 4(y - 1)^2 = 4y^2 - 12y + 9 - 4y^2 + 8y - 4 =5 - 4y; tag 6$



                thus,



                $x = dfrac{- 2y + 3 pm sqrt{5 - 4y}}{2(y - 1)^2}. tag 7$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Or, one might try this:



                  from



                  $x - sqrt x = yx + 1, tag 1$



                  we may write



                  $sqrt x= x - yx - 1; tag 2$



                  we square:



                  $ x = x^2 + y^2x^2 + 1 - 2yx^2 - 2x + 2yx$
                  $= (y^2 - 2y + 1)x^2 + (2y - 2)x + 1;tag 3$



                  subtract $x$:



                  $(y^2 - 2y + 1)x^2 + (2y - 3)x + 1 = 0,tag 3$



                  or



                  $(y - 1)^2 x^2 + (2y - 3)x + 1 = 0; tag 4$



                  now apply the quadratic formula:



                  $x = dfrac{-(2y - 3) pm sqrt{(2y - 3)^2 - 4(y - 1)^2}}{2(y - 1)^2}; tag 5$



                  simplify the radical:



                  $(2y - 3)^2 - 4(y - 1)^2 = 4y^2 - 12y + 9 - 4y^2 + 8y - 4 =5 - 4y; tag 6$



                  thus,



                  $x = dfrac{- 2y + 3 pm sqrt{5 - 4y}}{2(y - 1)^2}. tag 7$






                  share|cite|improve this answer









                  $endgroup$



                  Or, one might try this:



                  from



                  $x - sqrt x = yx + 1, tag 1$



                  we may write



                  $sqrt x= x - yx - 1; tag 2$



                  we square:



                  $ x = x^2 + y^2x^2 + 1 - 2yx^2 - 2x + 2yx$
                  $= (y^2 - 2y + 1)x^2 + (2y - 2)x + 1;tag 3$



                  subtract $x$:



                  $(y^2 - 2y + 1)x^2 + (2y - 3)x + 1 = 0,tag 3$



                  or



                  $(y - 1)^2 x^2 + (2y - 3)x + 1 = 0; tag 4$



                  now apply the quadratic formula:



                  $x = dfrac{-(2y - 3) pm sqrt{(2y - 3)^2 - 4(y - 1)^2}}{2(y - 1)^2}; tag 5$



                  simplify the radical:



                  $(2y - 3)^2 - 4(y - 1)^2 = 4y^2 - 12y + 9 - 4y^2 + 8y - 4 =5 - 4y; tag 6$



                  thus,



                  $x = dfrac{- 2y + 3 pm sqrt{5 - 4y}}{2(y - 1)^2}. tag 7$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  Robert LewisRobert Lewis

                  50.3k23369




                  50.3k23369






























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