Michael J. answered • 05/15/16
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Subtract H on both sides of the equation to isolate the t terms.
h - H = 16t2 - Vt
Divide both sides of the equation by 16. This will allow us to get a coefficient of 1 for t2.
(1/16)h - (1/16)H = t2 - (1/16)Vt
Next, we will complete the square. Divide the coefficient of t by 2 then square the result. We will add both sides of the equation by the value.
(-1/16)V / 2 = (-V/32)
(-V/32)2 = V2/1024
So now you have
(1/16)h - (1/16)H + (V/1024) = t2 - (1/16)Vt + (V2/1024)
(1/16)h - (1/16)H + (V/1024) = (t - (1/32)V)(t - (1/32)V)
(1/16)h - (1/16)H + (V/1024) = (t - (1/32)V)2
Now, we square-root both sides of the equation.
±√[(1/16)(h - H - (V/64))] =t - V/32
Add V/32 on both sides of the equation.
V/32 ± (1/4)√[(h - H - (V/64)] = t
V/32 ± (8/32)√[(h - H - (V/64)] = t
(V ± 8√[h - H - (V/64)]) / 32 = t
The final form of this solution looks similar to a solution when using the quadratic formula.
Michael J.
You could to deal with that rational term in the square-root part. The least common denominator in the square-root part is 1/64. You would use that to write an equivalent expression. Then use the rules of radicals. For example, the square-root part would be
√[(64h - 64H - V) / 64] =
√(64h - 64H - V) / √64 =
√(64h - 64H - V) / 8
Then you substitute that with the radical part I gave you.
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