Question

The biquadratic equation, two of whose roots are \[1+i\], \[1-\sqrt{2}\] is
A. \[{{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-2x-2=0\]
B. \[{{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}+2x+2=0\]
C. \[{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+2x-2=0\]
D. \[{{x}^{4}}+4{{x}^{3}}+5{{x}^{2}}-2x-2=0\]

Answer 1

Hint: A biquadratic equation is an equation that has 4 roots. In the given problem two roots of the biquadratic equation are given in which one of the roots is imaginary and the other one is irrational. The coefficients of the equation are real and rational, we will find the other two roots of the equation by solving the conjugates of the given roots. We have to take the conjugates as $1-i,1+\sqrt{2}$ . Then, we can express the roots as \[\left[ x-\left( 1+i \right) \right]\text{ }and\text{ }\left[ x-\left( 1-i \right) \right]\] , then multiply them and equate to 0 to get one equation. Similarly, we will find another equation for the next set of roots. Finally, we will multiply the obtained equations again to get the final answer.

Complete step by step answer:
Given two roots of the biquadratic equation are \[1+i\], \[1-\sqrt{2}\].
We can find the other two roots of the biquadratic equation by using the conjugates of the given roots.
To find the conjugate of complex numbers, we have to change the sign of the imaginary part.
So, we will get the conjugate of \[1+i=1-i\].
Similarly, the conjugate of an irrational number is calculated by changing the sign of the irrational part.
So, we will get the conjugate of \[1-\sqrt{2}=1+\sqrt{2}\].
Now, we have calculated all the 4 roots of the biquadratic equation.
All the four roots of the biquadratic equation are \[1+i\], \[1-\sqrt{2}\], \[1-i\], \[1+\sqrt{2}\].
We know that if we have two roots of an equation as a and b, then we can express the equation as (x-a) (x-b) = 0.
Using this, we will find the quadratic equation with two imaginary roots \[1+i\] and \[1-i\], which is given as
\[\begin{align}
  & \left[ x-\left( 1-i \right) \right]\left[ x-\left( 1+i \right) \right]=0 \\
 & \left[ x-1+i \right]\left[ x-1-i \right]=0 \\
 & \left[ (x-1)+i \right]\left[ (x-1)-i \right]=0 \\
\end{align}\]
Now using identity $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ , we have
\[\begin{align}
  & {{\left( x-1 \right)}^{2}}-\left( {{i}^{2}} \right)=0 \\
 & \because {{i}^{2}}=-1 \\
 & {{\left( x-1 \right)}^{2}}+1=0 \\
\end{align}\]
Now we will form the quadratic equation with irrational roots \[1-\sqrt{2}\] and \[1+\sqrt{2}\], which is
\[\begin{align}
  & \left[ x-\left( 1-\sqrt{2} \right) \right]\left[ x-\left( 1+\sqrt{2} \right) \right]=0 \\
 & \left[ x-1+\sqrt{2} \right]\left[ x-1-\sqrt{2} \right]=0 \\
 & \left[ \left( x-1 \right)+\sqrt{2} \right]\left[ \left( x-1 \right)-\sqrt{2} \right]=0 \\
\end{align}\]
Now using identity $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ , we have
\[\begin{align}
  & {{\left( x-1 \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}=0 \\
 & {{\left( x-1 \right)}^{2}}-2=0 \\
\end{align}\]
We will now find the product of these two quadratic equations to get the required biquadratic equation
\[\begin{align}
  & \left[ {{\left( x-1 \right)}^{2}}+1 \right]\left[ {{\left( x-1 \right)}^{2}}-2 \right]=0 \\
 & {{\left( x-1 \right)}^{4}}+{{\left( x-1 \right)}^{2}}-2{{\left( x-1 \right)}^{2}}-2=0 \\
 & {{\left( x-1 \right)}^{4}}-{{\left( x-1 \right)}^{2}}-2=0 \\
\end{align}\]
Now, expanding the terms in the above quadratic equations using \[{{\left( a-b \right)}^{4}}={{a}^{4}}-4{{a}^{3}}\text{b}+6{{a}^{2}}{{b}^{2}}-4a{{b}^{3}}+{{b}^{4}}\text{ and }{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] , we will get
$\begin{align}
  & \left( {{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}-4x+1 \right)-\left( {{x}^{2}}-2x+1 \right)-2=0 \\
 & {{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}-4x+1-{{x}^{2}}+2x-1-2=0 \\
 & {{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-2x-2=0 \\
\end{align}$
\[{{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-2x-2=0\]
Hence the required biquadratic equation is \[{{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-2x-2=0\], option A is the correct one.

Note:
Avoid silly mistakes in solving the conjugate of the given imaginary and irrational roots of the biquadratic equation. Be clear about forming the quadratic equation with the given roots and also a biquadratic equation is formed by the product of two quadratic equations. There is a possibility that some students might try to substitute the roots in the given options and check if it satisfies or not. This is not at all recommended as it is very time consuming and we are supposed to check if all the four roots \[1+i\], \[1-\sqrt{2}\], \[1-i\], \[1+\sqrt{2}\] satisfy the equations or not.