Question

If $\frac{4x^2+3y^2}{4x^2-3y^2}=\frac{11}{5}$ then $x^2=2y^2$.

• A. True
• B. False

Answer 1

The correct option is A True

Given:

$\frac{4x^2+3y^2}{4x^2-3y^2}=\frac{11}{5}$

Using componendo and dividendo:

$\frac{4x^2+3y^2+4x^2-3y^2}{4x^2+3y^2-4x^2+3y^2}=\frac{11+5}{11-5}$

$⟹\frac{8x^2}{6y^2}=\frac{16}{6}$

$⟹x^2=2y^2$