Question

-3x^4+27x^2+1200=0 Please show all steps

Answer 1

Answer:

x = 5, -5, 4i, -4i   ; $\mathbf{\sqrt{-1}=i}$

Step-by-step explanation:

$3\textrm{x}^{4}-27\textrm{x}^{2}-1200=0$

assume $\textrm{x}^{2}=\textrm{t}$

$3\textrm{t}^{2}-27\textrm{t}-1200=0$

Now the above equation is a quadratic equation.

There are two solutions of any quadratic equation. Solution of a quadratic equation $\mathbf{a\textrm{x}^{2}+b\textrm{x}+c=0}$ is given by:

$\mathbf{\textrm{x}=\frac{-b+\sqrt{b^{2}-4ac}}{2a},\textrm{x}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$

similarly there are two solutions of the quadratic equation $3\textrm{t}^{2}-27\textrm{t}-1200=0$ which are:

$\textrm{t}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}=\frac{-(-27)+\sqrt{(-27)^{2}-4 \cdot 3 \cdot (-1200)}}{2 \cdot 3}=25,\\ \textrm{t}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{-(-27)-\sqrt{(-27)^{2}-4 \cdot 3 \cdot (-1200)}}{2 \cdot 3}=-16$

Since $\textrm{x}^{2}=\textrm{t}$

Therefore $\textrm{x}^{2}=25,\textrm{x}^{2}=-16$

$\textrm{x}^{2}=25 \Rightarrow \textrm{x}=+\sqrt{25},-\sqrt{25} \Rightarrow \textrm{x}=5,-5$

$\textrm{x}^{2}=-16 \Rightarrow \textrm{x}=+\sqrt{-16},-\sqrt{-16} \Rightarrow \textrm{x}=\sqrt{-1} \cdot \sqrt{16},-\sqrt{-1} \cdot \sqrt{16} \Rightarrow \textrm{x}=4i,-4i$ ; where $\textrm{i}=\sqrt{-1}$ (the numbers with 'i' are called imaginary numbers)

Therefore $\mathbf{x=5,-5,4i,-4i}$