How many roots, real or complex, does the polynomial $7 + 5 x^{4} - 3 x^{2} = 0$ have in all?

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Solution

Explanation for the correct answer:
Step 1: Writing the given equation in quadratic form.
We have been given $7 + 5 x^{4} - 3 x^{2} = 0$
On solving, we get:
$\begin{array}{rcl} 7 + 5 x^{4} - 3 x^{2} & = & 0 \end{array}$
Using quadratic equation roots formula as if $a x^{2} + b x + c = 0$ then
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Here, for equation $\begin{array}{rcl} 7 + 5 x^{4} - 3 x^{2} & = & 0 \end{array}$ consider $x^{2} = X$ then,
Step 2: Solving Quadratic equation to find roots.
$\begin{array}{rcl} 7 + 5 X^{2} - 3 X & = & 0 \end{array}$ so, we get
$\begin{array}{rcl} \Rightarrow & X = \frac{3 \pm \sqrt{9 - 4 (5) (7)}}{2 (5)} \\ \Rightarrow & X = \frac{3 \pm \sqrt{- 131}}{10} \\ \Rightarrow & x = \pm \sqrt{\frac{3 + \sqrt{- 131}}{10}}, \pm \sqrt{\frac{3 - \sqrt{- 131}}{10}} \end{array}$
Therefore, the given polynomial has four complex roots.

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