Chapter 6.7, Problem 26PPS

To determine

Solve the given equation. State the number and type of roots.

Answer to Problem 26PPS

The above equation has five roots, four are imaginary $i,-i,i,-i$ , one is real 0.

Explanation of Solution

Given information:

The equation $x^5+2x^3+x=0$ .

Formula used:

Splitting the middle term: To solve a quadratic equation $a{x}^2+bx+c=0$ , the middle term of equation is splitted in such a way that the product of coefficient of its parts is equal to product of coefficient of first and third term.

  $x^5+2x^3+x=0$

Taking $x$ common from LHS,

  $x\left(x^4+2x^2+1\right)=0$

In above equation, either one or two of $x^4+x^2+1$ and $x$ is zero,

Factorizing $x^4+2x^2+1$

  $x^4+2x^2+1$

Let $x^2=z$

Recall that in splitting the middle term, the middle term of equation is splitted in such a way that the product of coefficient of its parts is equal to product of coefficient of first and third term.

  $\begin{array}{l}{x}^4+2x^2+1\\ \\ =z^2+2z+1\\ \\ =z^2+z+z+1\\ \\ =\left(z+1\right)\left(z+1\right)\end{array}$

As $x^2=z$

  $=\left(x^2+1\right)\left(x^2+1\right)$

  $\begin{array}{l}{x}^2+1=0\\ \\ x^2=-1\\ \\ x=\pm i\end{array}$

  $\begin{array}{l}{x}^2+1=0\\ \\ x^2=-1\\ \\ x=\pm i\end{array}$

The above equation has five roots, four are imaginary $i,-i,i,-i$ , one is real 0.