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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

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

edited 5 hours ago

Robert Lewis

50.3k23369

asked 8 hours ago

TriangleTriangle

854

$endgroup$

• 
  
  
  

  
  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
  
  
  

  
  
  
  

add a comment | 

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

edited 5 hours ago

Robert Lewis

50.3k23369

asked 8 hours ago

TriangleTriangle

854

$endgroup$

• 
  
  
  

  
  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
  
  
  

  
  
  
  

add a comment | 

$begingroup$

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

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

algebra-precalculus

share|cite|improve this question

edited 5 hours ago

Robert Lewis

50.3k23369

asked 8 hours ago

TriangleTriangle

854

$endgroup$

share|cite|improve this question

edited 5 hours ago

Robert Lewis

50.3k23369

asked 8 hours ago

TriangleTriangle

854

share|cite|improve this question

edited 5 hours ago

Robert Lewis

50.3k23369

asked 8 hours ago

TriangleTriangle

854

edited 5 hours ago

Robert Lewis

50.3k23369

edited 5 hours ago

Robert Lewis

50.3k23369

asked 8 hours ago

TriangleTriangle

854

asked 8 hours ago

TriangleTriangle

854

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

answered 7 hours ago

DunkelDunkel

325111

$endgroup$

add a comment | 

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

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369

$endgroup$

add a comment | 

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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

answered 7 hours ago

DunkelDunkel

325111

$endgroup$

add a comment | 

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

answered 7 hours ago

DunkelDunkel

325111

$endgroup$

add a comment | 

$begingroup$

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

answered 7 hours ago

DunkelDunkel

325111

$endgroup$

share|cite|improve this answer

answered 7 hours ago

DunkelDunkel

325111

answered 7 hours ago

DunkelDunkel

325111

answered 7 hours ago

DunkelDunkel

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

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369

$endgroup$

add a comment | 

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

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369

$endgroup$

add a comment | 

$begingroup$

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

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369

answered 6 hours ago

Robert LewisRobert Lewis

50.3k23369