Question

Solve the equation ${{x}^{4}}-4{{x}^{2}}+8x+35=0$ having given that one root is $2+\sqrt{-3}$
A. The roots of the given equation are $-2\pm i,2\pm \sqrt{3}i$
B. The roots of the given equation are $2\pm i,2\pm \sqrt{3}i$
C. The roots of the given equation are $\pm 2+i,2\pm \sqrt{3}i$
D. The roots of the given equation are $\pm 2-i,2\pm \sqrt{3}i$

Answer 1

Here we have been given one biquadratic equation and one factor which is imaginary. We will here make use of the fact that imaginary roots are conjugate of each other. Hence, the second root of this equation will be the conjugate of the given root. Then, we will multiply both the obtained factors to get a quadratic factor, and then we will divide the given equation by the now obtained quadratic factor. As a result we will obtain another quadratic factor of the equation and we will obtain the roots of that by using the quadratic formula for any quadratic equation $a{{x}^{2}}+bx+c=0$ given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. As a result, we will obtain the remaining factors. Hence, we will obtain all the required factors and thus the required answers.

Complete step-by-step solution:
Here, we have been given the equation ${{x}^{4}}-4{{x}^{2}}+8x+35=0$ and we have been given that $2+\sqrt{-3}$ is one of the roots of this equation.
Now, we know that $\sqrt{-1}=i$
Hence, we get the root as:
$\begin{aligned} & 2+\sqrt{-3} \\\ & \Rightarrow 2+\sqrt{-1}\sqrt{3} \\\ & \Rightarrow 2+\sqrt{3}i \\\ \end{aligned}$
Now, we can see that this is an imaginary root.
We know that if ‘z’ being a complex number is a root of any equation, then its conjugate, i.e. $\bar{z}$ is also a root of the same equation.
Thus, we get the second root of the given equation as:
$\begin{aligned} & \overline{2+\sqrt{3}i} \\\ & \Rightarrow 2-\sqrt{3}i \\\ \end{aligned}$
Now, we have two roots of the equation as:
$\begin{aligned} & x=2+\sqrt{3}i \\\ & x=2-\sqrt{3}i \\\ \end{aligned}$
Thus, we can say that $x-2-\sqrt{3}i=0$ and $x-2+\sqrt{3}i=0$ are the factors of the given equation.
Now, we know that if $x-a$ and $x-b$ are the factors of any polynomial, then $\left( x-a \right)\left( x-b \right)$ are also the factors of the same equation.
Thus, we can say that $\left( x-2-\sqrt{3}i \right)\left( x-2+\sqrt{3}i \right)$ is a factor of the given equation.
On solving, we get this factor as:
$\begin{aligned} & \left( x-2-\sqrt{3}i \right)\left( x-2+\sqrt{3}i \right) \\\ & \Rightarrow {{\left( x-2 \right)}^{2}}-{{\left( \sqrt{3}i \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}-4x+4-\left( -3 \right) \\\ & \Rightarrow {{x}^{2}}-4x+7 \\\ \end{aligned}$
Now, we will divide the now obtained quadratic factor by the given equation and hence we will obtain the remaining factors.
Dividing ${{x}^{4}}-4{{x}^{2}}+8x+35=0$ by ${{x}^{2}}-4x+7$, we get:
${{x}^{2}}-4x+7\overset{{{x}^{2}}+4x+5}{\overline{\left){\begin{aligned} & {{x}^{4}}-4{{x}^{2}}+8x+35 \\\ & \underline{-\left( {{x}^{4}}-4{{x}^{3}}+7{{x}^{2}} \right)} \\\ & 4{{x}^{3}}-11{{x}^{2}}+8x+35 \\\ & \underline{-\left( 4{{x}^{3}}-16{{x}^{2}}+28 \right)} \\\ & 5{{x}^{2}}-20x+35 \\\ & \underline{-\left( 5{{x}^{2}}-20x+35 \right)} \\\ & 0 \\\ \end{aligned}}\right.}}$
Thus, the other factor of this equation is ${{x}^{2}}+4x+5=0$
Now, we will find the other factors of this equation by using the quadratic formula for any quadratic equation $a{{x}^{2}}+bx+c=0$ given as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here,
$\begin{aligned} & a=1 \\\ & b=4 \\\ & c=5 \\\ \end{aligned}$
Now, putting these values in the above mentioned formula we get:
$\begin{aligned} & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( 5 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{16-20}}{2} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{-4}}{2} \\\ \end{aligned}$
Now, we can write $\sqrt{-4}$ as $\sqrt{-4}=\sqrt{-1}\sqrt{4}=\sqrt{4}i$
Thus, we get:
$\begin{aligned} & x=\dfrac{-4\pm \sqrt{-4}}{2} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{4}i}{2} \\\ & \Rightarrow x=\dfrac{-4\pm 2i}{2} \\\ & \therefore x=-2\pm i \\\ \end{aligned}$
Hence, the remaining factors of the given equation except for the given factor are $2-\sqrt{3}i,-2\pm i$
Thus, the factors of the given equation are $-2\pm i,2\pm \sqrt{3}i$
Hence, option (A) is the correct option.

Note: Always remember, that in any polynomial of any degree, if there is one imaginary factor, then there is also a second imaginary factor to this equation which is the conjugate of the first equation.
For example:
1. In any quadratic equation there are two roots. Hence, there can be either two real roots or two imaginary roots which are the conjugate of each other.
2. In any cubic equation there are 3 roots. Hence, there can be either 3 real roots or one real and 2 imaginary roots which are the conjugate of each other.
Thus, we can also say that imaginary roots always occur in pairs.