Question

-3x^4+27x^2+1200=0

Please show all steps

Answer 1

x = 5, -5, 4i, -4i ;
`\mathbf{√(-1)=i}`

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}=+√(25),-√(25) \Rightarrow \textrm{x}=5,-5`

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

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