Question

Find the real roots of the equation
36÷x^4+4=25÷x2​

Answer 1

Answer: x = ±$\frac{3}{2}$,   x = ±2

Explanation:

$\frac{36}{x^4} + 4 = \frac{25}{x^2}$

Multiply $x^4$ on both sides

$x^4(\frac{36}{x^4} + 4) = x^4(\frac{25}{x^2})$

$\frac{36x^4}{x^4} + 4x^4 = \frac{25x^4}{x^2}$

Simplify the fractions

$36 + 4x^4 = 25x^2$

$4x^4 - 25x^2 + 36 = 0$

Let y = $x^2$

$4y^2 - 25y + 36 = 0$

Now solve as a normal quadratic equation

$4y^2 -16y - 9y + 36 = 0$

$4y(y-4) - 9(y-4) = 0$

$4y - 9 = 0$ (or) $y - 4 = 0$

$y = \frac{9}{4}$ (or) $y = 4$

Now use these values to get the values of x\

$x^{2} = y$

$x^{2} = \frac{9}{4}$ (or) $x^{2} = 4$

$x = \sqrt{\frac{9}{4} }$ (or) $x = \sqrt{4}$

x = ±$\frac{3}{2}$ (or) x = ±2