1 Answer

Ishaaq · Answer 1 · 2021-05-06T23:08:19+0000

Answer:

c = ±
$\sqrt{(d+dt^2)/(t^2-3) }$

Explanation:

Given

t =
$\sqrt{(3c^2+d)/(c^2-d) }$ ( square both sides )

t² =
(3c^2+d)/(c^2-d) ( multiply both sides by c² - d )

t²(c² - d) = 3c² + d ← distribute left side

t²c² - dt² = 3c² + d ( subtract 3c² from both sides )

t²c² - 3c² - dt² = d ( add dt² to both sides )

t²c² - 3c² = d +dt² ← factor out c² from each term on the left side

c²(t² - 3) = d + dt² ( divide both sides by t² - 3 )

c² =
(d+dt^2)/(t^2-3) ( take the square root of both sides )

c = ±
$\sqrt{(d+dt^2)/(t^2-3) }$

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