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I have this equation \(c=\frac{2at-{t}^{2}b}{2}\) and I need to solve for t but I'm having trouble.

 

Here is what I've done so far:

 

1. multiply both sides by 2.

 

2c = 2at - t²b

 

2. factor out t.

 

2c = t(2a-tb)

 

3. divide both sides by t.

 

2c / t = (2a-tb)

 

4. switcheroo

 

2c / (2a-tb) = t

 

That didn't uniquely solve for t so I tried another approach starting from step 2.

 

2. Add t²b to both sides.

 

2c + t²b = 2at

 

3. subtract 2c from both sides.

 

t²b = 2at - 2c

 

4. divide both sides by b.

 

t² = (2at - 2c) / b

 

5. square root both sides.

 

\(t=±\sqrt{\frac{2at-2c}{b}}\)

 

This also didn't uniquely solve for t, and I got very suspicious that it might not be uniquely solvable for t, even though I already solved it for a and b. Is it solvable, and if so, what would it be?

 

I realize that solving for one variable among 4 is probably pretty difficult, but I don't know exactly how difficult it would be to solve this equation. I'd appreciate any help though.

 Nov 12, 2016

Best Answer 

 #4
avatar+128079 
+5

[2at - t^2b] / 2  = c      multiply both sides by 2

 

2at - t^2b  = 2c    multiply throgh by -1

 

bt^2 - 2at   =  -2c         we will complete the square on t.......divide both sides by b

 

t^2 - (2a/b)*t  = [-2c] /b       take 1/2 the coefficient on  t  =  a/b   .....square it and add to both sides

 

t^2  - (2a/b)*t  + (a/b)^2  =  (a/b)^2 - [2c]/b    factor the left side

 

[ t - (a/b) ]^2  =   (a/b)^2 - [2c]/b      take the pos/neg roots of both sides

 

t - (a/b)    =  ± √ [  (a/b)^2 - [2c]/b]     add (a/b) to both sides

 

t =  ± √ [  (a/b)^2 - [2c]/b]   + (a/b)  we can clean this up a little

 

t = ± √ [ a^2 - 2bc] / b + a/b  

 

t = ( a  ± √ [ a^2 - 2bc] )  / b       [ we assume that b is not  0 ]

 

 

 

cool cool cool

 Nov 12, 2016
 #1
avatar
+5

Solve for t:
c = 1/2 (2 a t-b t^2)

c = 1/2 (2 a t-b t^2) is equivalent to 1/2 (2 a t-b t^2) = c:
1/2 (2 a t-b t^2) = c

Multiply both sides by 2:
2 a t-b t^2 = 2 c

Divide both sides by -b:
t^2-(2 a t)/b = -(2 c)/b

Add a^2/b^2 to both sides:
a^2/b^2-(2 a t)/b+t^2 = a^2/b^2-(2 c)/b

Write the left hand side as a square:
(t-a/b)^2 = a^2/b^2-(2 c)/b

Take the square root of both sides:
t-a/b = sqrt(a^2/b^2-(2 c)/b) or t-a/b = -sqrt(a^2/b^2-(2 c)/b)

Add a/b to both sides:
t = a/b+sqrt(a^2/b^2-(2 c)/b) or t-a/b = -sqrt(a^2/b^2-(2 c)/b)

Add a/b to both sides:
Answer: |t = a/b+sqrt(a^2/b^2 - (2c)/b)      or        t = a/b-sqrt(a^2/b^2 - (2c)/b)

 Nov 12, 2016
 #2
avatar+495 
0

Thank you so much, guest!

 Nov 12, 2016
 #3
avatar
0

You are most welcome.

 Nov 12, 2016
 #4
avatar+128079 
+5
Best Answer

[2at - t^2b] / 2  = c      multiply both sides by 2

 

2at - t^2b  = 2c    multiply throgh by -1

 

bt^2 - 2at   =  -2c         we will complete the square on t.......divide both sides by b

 

t^2 - (2a/b)*t  = [-2c] /b       take 1/2 the coefficient on  t  =  a/b   .....square it and add to both sides

 

t^2  - (2a/b)*t  + (a/b)^2  =  (a/b)^2 - [2c]/b    factor the left side

 

[ t - (a/b) ]^2  =   (a/b)^2 - [2c]/b      take the pos/neg roots of both sides

 

t - (a/b)    =  ± √ [  (a/b)^2 - [2c]/b]     add (a/b) to both sides

 

t =  ± √ [  (a/b)^2 - [2c]/b]   + (a/b)  we can clean this up a little

 

t = ± √ [ a^2 - 2bc] / b + a/b  

 

t = ( a  ± √ [ a^2 - 2bc] )  / b       [ we assume that b is not  0 ]

 

 

 

cool cool cool

CPhill Nov 12, 2016
 #5
avatar+495 
+5

Thanks so much, CPhill! I forgot about completing the square. I appreciate your response!

 Nov 12, 2016

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