Question

Find all the zeroes of the equation(with simple steps).

Answer 1

Answer-

The zeros are, $5,\ -5,\ 4i,\ -4i$

Solution-

$\Rightarrow -3x^4+27x^2+1200=0$

$\Rightarrow -3(x^2)^2+27(x^2)+1200=0$

Here,

a = -3, b = 27, c = 1200

So,

$x^2=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}$

$=\dfrac{-27\pm \sqrt{729+14400}}{-6}$

$=\dfrac{-27\pm 123}{-6}$

$=\dfrac{-27+123}{-6},\ \dfrac{-27- 123}{-6}$

$=\dfrac{96}{-6},\ \dfrac{-150}{-6}$

$=-16,\ 25$

So,

$\Rightarrow x^2=25,\ -16$

$\Rightarrow x=\sqrt{25},\ \sqrt{-16}$

$\Rightarrow x=5,\ -5,\ 4i,\ -4i$