Question

-3x^4+27x^2+1200=0Find all the zeros of the equation y problem is the calculation after the info in the picture

Answer 1

`\begin{gathered} \text{Given} \\ -3x^4+27x^2+1200=0 \end{gathered}`

Transform the variable x² into u

`-3x^4+27x^2+1200=0\Longrightarrow-3u^2+27u+1200=0`

Use the quadratic formula to solve for solutions in u, with a = -3, b = 27, c = 1200.

`\begin{gathered} u=( -b \pm√(b^2 - 4ac))/( 2a ) \\ u=( -27 \pm√(27^2 - 4(-3)(1200)))/( 2(-3) ) \\ u=\frac{-27\pm\sqrt[]{729-(-14400)}}{-6} \\ u=( -27 \pm√(15129))/( -6 ) \\ u=( -27 \pm123\, )/( -6 ) \\ \\ u_1=(-27+123)/(-6)=(96)/(-6)=-16 \\ u_2=(-27-123)/(-6)=(-150)/(-6)=25 \\ \\ \text{Which means that if we factor} \\ -3u^2+27u+1200=0 \\ \text{Then it is} \\ (u+16)(u-25)=0 \end{gathered}`

Revert u back into x²

`\begin{gathered} (u+16)(u-25)=0\Longrightarrow(x^2+16)(x^2-25)=0 \\ \\ \text{Recall the special factor} \\ (a^2-b^2)=(a+b)(a-b)\text{ and apply it to }(x^2-25) \\ \\ (x^2+16)(x^2-25)=0 \\ (x^2+16)(x+5)(x-5)=0 \end{gathered}`

Equate to zero all the factors, and solve for x

`\begin{gathered} x^2+16=0 \\ x^2=-16 \\ \sqrt[]{x^2}=-16 \\ x_1=\pm\sqrt[]{-16} \\ \\ x+5=0 \\ x_2=-5 \\ \\ x-5=0 \\ x_3=5 \end{gathered}`

The real zeroes to the given equation is x = -5, and x = 5.