Chapter 2.5, Problem 23E

To determine

Find the zeros of the function $f(x)=x^2+25$ and write the polynomial as a product of linear factors and verify zeros using graphical method.

Answer to Problem 23E

  $5i,-5i$

Explanation of Solution

Given:

Function: $f(x)=x^2+25$

Formula used:

The zeros of a quadratic equation $a{x}^2+bx+c=0$ can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\Rightarrow f(x)=x^2+25$

To find zeros, put $f(x)=0,$

  $\Rightarrow x^2+25=0$

Applying above formula,

  $\begin{array}{l}\Rightarrow x=\frac{-0\pm \sqrt{{(0)}^2-4(1)(25)}}{2(1)}\\ \Rightarrow x=\frac{0\pm \sqrt{0-100}}{2}\\ \Rightarrow x=\frac{\pm \sqrt{-100}}{2}\\ \Rightarrow x=\frac{\pm 10\sqrt{-1}}{2}\\ \Rightarrow x=\pm 5i\\ \Rightarrow x=5i,-5i\end{array}$

Now, writing the given function as product of linear factors,

  $\begin{array}{l}\Rightarrow f(x)=(x-(5i))(x-(-5i))\\ \Rightarrow f(x)=(x-5i)(x+5i)\end{array}$

As the zeros of the function are imaginary zeros, verification of roots through graphical method is not possible.

Conclusion:

Therefore, the zeros of given function are $x=-5i\text{and}5i.$