Question

How do you solve the quadratic using the quadratic formula given ${{x}^{4}}+16{{x}^{2}}-225=0$ over the set of complex numbers?

Answer 1

First convert the given equation to the form for which the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ will be applicable. Do the necessary calculations to find the values of ${{x}^{2}}$. Again using those values find the values of ‘x’.

Complete step-by-step solution:
Solving the equation means we have to get every value (roots) of ‘x’ for which the equation gets satisfied.
Solving the quadratic equation: If we have an equation of the form $a{{x}^{2}}+bx+c=0$, then we can solve the equation using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Considering our equation ${{x}^{4}}+16{{x}^{2}}-225=0$
It can be written as ${{\left( {{x}^{2}} \right)}^{2}}+16\left( {{x}^{2}} \right)-225=0$
Now, it is in the form $a{{x}^{2}}+bx+c=0$ for which quadratic formula is applicable.
By comparison $a=1$, $b=16$ and $c=-225$
So, ${{x}^{2}}$ can be determined using the quadratic formula as

& {{x}^{2}}=\dfrac{-16\pm \sqrt{{{\left( 16 \right)}^{2}}-4\cdot 1\cdot \left( -225 \right)}}{2\cdot 1} \\\ & \Rightarrow {{x}^{2}}=\dfrac{-16\pm \sqrt{256-\left( - 900 \right)}}{2} \\\ & \Rightarrow {{x}^{2}}=\dfrac{-16\pm \sqrt{256+900}}{2} \\\ & \Rightarrow {{x}^{2}}=\dfrac{-16\pm \sqrt{1156}}{2} \\\ & \Rightarrow {{x}^{2}}=\dfrac{-16\pm 34}{2} \\\ & \Rightarrow {{x}^{2}}=-8\pm 17 \\\ \end{aligned}$$ Either $\begin{aligned} & {{x}^{2}}=-8+17 \\\ & \Rightarrow {{x}^{2}}=9 \\\ & \Rightarrow x=\sqrt{9} \\\ & \Rightarrow x=\pm 3 \\\ \end{aligned}$ Or, $\begin{aligned} & {{x}^{2}}=-8-17 \\\ & \Rightarrow {{x}^{2}}=-25 \\\ & \Rightarrow x=\sqrt{-25} \\\ & \Rightarrow x=\pm 5i \\\ \end{aligned}$ **So, the solution of the equation ${{x}^{4}}+16{{x}^{2}}-225=0$ is $x=\pm 3\text{ and }\pm 5i$, This is the required solution of the given question.** **Note:** Bringing the given equation to quadratic form should be the first approach for solving this question. We encountered $\sqrt{-25}$ during the calculation. As we know $i=\sqrt{-1}$, so $\sqrt{-25}$ can be simplified as $\sqrt{-25}=\sqrt{25\cdot \left( -1 \right)}=\sqrt{25}\cdot \sqrt{\left( -1 \right)}=5i$. Since the degree of the given equation is 4, so we are getting 4 different values of ‘x’ as the roots of the equation.