So, making x subject of the formula, x = [m - 2pt³ ±√(m² - 4pt²m)]/{2t⁵}
How to make x subject of the formula?
Since p = √(mx/t) - t²x
So, p + t²x = √(mx/t)
Squaring both sides, we have
(p + t²x)² = [√(mx/t)]²
p² + 2pt²x + t⁴x² = mx/t
Multiplying through by t,we have
(p² + 2pt²x + t⁴x²)t = mx/t × t
p²t + 2pt³x + t⁵x² = mx
p²t + 2pt³x + t⁵x² - mx = 0
t⁵x² + 2pt³x - mx + p²t = 0
t⁵x² + (2pt³ - m)x + p²t = 0
Using the quadratic formula, we find x.
[tex]x = \frac{-b +/-\sqrt{b^{2} - 4ac} }{2a}[/tex]
where
- a = t⁵,
- b = (2pt³ - m) and
- c = p²t
Substituting the values of the variables into the equation, we have
[tex]x = \frac{-(2pt^{3} - m) +/-\sqrt{(2pt^{3} - m)^{2} - 4(t^{5})(p^{2}t) } }{2t^{5} }\\= \frac{-(2pt^{3} - m) +/-\sqrt{4p^{2} t^{6} - 4pt^{2}m + m^{2} - 4p^{2}t^{6} } }{2t^{5}}\\= \frac{-(2pt^{3} - m) +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}\\= \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}[/tex]
So, making x subject of the formula, [tex]x = \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}[/tex]
Learn more about subject of formula here:
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