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Ivahew [28]
3 years ago
6

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

Mathematics
1 answer:
sveta [45]3 years ago
6 0

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}

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