Question

Pls help to solve this qns ASAP with step by step explanation. Thanks!

Answer 1

Answer:

c = ± $\sqrt{\frac{d+dt^2}{t^2-3} }$

Step-by-step explanation:

Given

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

t² = $\frac{3c^2+d}{c^2-d}$ ( 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 = $\frac{d+dt^2}{t^2-3}$ ( take the square root of both sides )

c = ± $\sqrt{\frac{d+dt^2}{t^2-3} }$