+495 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.
+129899 [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 ]
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)
+129899 [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 ]
