Respuesta :
Answer:
x = 5, -5, 4i, -4i ; [tex]\mathbf{\sqrt{-1}=i}[/tex]
Step-by-step explanation:
[tex]3\textrm{x}^{4}-27\textrm{x}^{2}-1200=0[/tex]
assume [tex]\textrm{x}^{2}=\textrm{t}[/tex]
[tex]3\textrm{t}^{2}-27\textrm{t}-1200=0[/tex]
Now the above equation is a quadratic equation.
There are two solutions of any quadratic equation. Solution of a quadratic equation [tex]\mathbf{a\textrm{x}^{2}+b\textrm{x}+c=0}[/tex] is given by:
[tex]\mathbf{\textrm{x}=\frac{-b+\sqrt{b^{2}-4ac}}{2a},\textrm{x}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}[/tex]
similarly there are two solutions of the quadratic equation [tex]3\textrm{t}^{2}-27\textrm{t}-1200=0[/tex] which are:
[tex]\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[/tex]
Since [tex]\textrm{x}^{2}=\textrm{t}[/tex]
Therefore [tex]\textrm{x}^{2}=25,\textrm{x}^{2}=-16[/tex]
[tex]\textrm{x}^{2}=25 \Rightarrow \textrm{x}=+\sqrt{25},-\sqrt{25} \Rightarrow \textrm{x}=5,-5[/tex]
[tex]\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[/tex] ; where [tex]\textrm{i}=\sqrt{-1}[/tex] (the numbers with 'i' are called imaginary numbers)
Therefore [tex]\mathbf{x=5,-5,4i,-4i}[/tex]
