Question

Solve the equation ${{x}^{4}}-8{{x}^{3}}+24{{x}^{2}}-32x+20=0$ if $3+i$ is a root of it.

Answer 1

Hint: We have been given the equation ${{x}^{4}}-8{{x}^{3}}+24{{x}^{2}}-32x+20=0$ . Now the one root is given the other root will be $3-i$ . After that, you will get the equation from these two roots. Then $({{x}^{2}}+px+2)$ is another factor, so after that compare the equation and you will get the other factor and roots of the equation. Try it, you will get the answer.

Complete step-by-step answer:
In Mathematics, an equation is a statement that asserts the equality of two expressions. Solving an equation containing variables consists of determining which values of the variables make the equality true. Variables are also called unknowns and the values of the unknowns that satisfy the equality are called solutions of the equation. There are two kinds of equations: identities and conditional equations. An identity is true for all values of the variable. A conditional equation is only true for particular values of the variables.
Single variable equation can be $x+2=0$.
An equation is written as two expressions, connected by an equals sign ("$=$"). The expressions on the two sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation.
The most common type of equation is an algebraic equation, in which the two sides are algebraic expressions. Each side of an algebraic equation will contain one or more terms. For example, the equation, $A{{x}^{2}}+Bx+C=y$has left-hand side $A{{x}^{2}}+Bx+C$ which has three terms, and right-hand side $y$, consisting of just one term. The unknowns are $x$ and $y$ and the parameters are $A,$ $B,$ and $C$.
We have been given that $3+i$ is the root of the equation, so the other root will be $3-i$ .
Now sum of roots $=3+i+3-i=6$ .
And product of roots $=(3+i)(3-i)=9-{{i}^{2}}=9+1=10$ . …….(${{i}^{2}}=-1$)
So the equation of $3+i$ and $3-i$ roots is,
${{x}^{2}}-6x+10=0$
${{x}^{4}}-8{{x}^{3}}+24{{x}^{2}}-32x+20=({{x}^{2}}-6x+10)({{x}^{2}}+px+2)$
Now equating coefficient of $x$ we get,
$\begin{align}
  & {{x}^{4}}-8{{x}^{3}}+24{{x}^{2}}-32x+20={{x}^{4}}+p{{x}^{3}}+2{{x}^{2}}-6{{x}^{3}}-6p{{x}^{2}}-12x+10{{x}^{2}}+10px+20 \\
 & {{x}^{4}}-8{{x}^{3}}+24{{x}^{2}}-32x+20={{x}^{4}}+(p-6){{x}^{3}}+(-6p+12){{x}^{2}}+(10p-12)x+20 \\
\end{align}$
Now we get,
$\begin{align}
  & -12+10p=-32 \\
 & 10p=-20 \\
 & p=-2 \\
\end{align}$
So the other factor is ${{x}^{2}}-2x+2$ ,
Therefore, now comparing above equation with $a{{x}^{2}}+bx+c=0$ we get,
$a=1,b=-2,c=2$ ,
We know that, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
$x=\dfrac{-(-2)\pm \sqrt{{{(-2)}^{2}}-4(1)(2)}}{2(1)}=\dfrac{2\pm \sqrt{4-8}}{2}=\dfrac{2\pm \sqrt{-4}}{2}=1\pm i$
The two factors of equation ${{x}^{4}}-8{{x}^{3}}+24{{x}^{2}}-32x+20=0$ are ${{x}^{2}}-6x+10$ and ${{x}^{2}}-2x+2$ .
Hence the equation is solved.

Note:The important point that we must remember is that the complex root always has its conjugate pair which will also be the root of the given equation. If the student is not aware of this, it might get difficult to solve this question. Then, we have to take the factor as $({{x}^{2}}+px+2)$ . We can see that the constant term of the given equation is 20 and we have a factor as ${{x}^{2}}-6x+10$ , so in order to get 20 as the constant term when both factors are multiplied, we must consider the last term of the second factor as 2. This is an easy way to remember how to express the second factor.