Chapter 2.5, Problem 11E

To determine

To verify: The function $f\left(x\right)=x^2+6$ has the indicated roots $-\sqrt{6}i,\sqrt{6}i$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=x^2+6$ and the roots $-\sqrt{6}i,\sqrt{6}i$ .

Formula used:

Complex zeros of polynomial functions are imaginary numbers that cannot be plotted or represented on the graph.

To prove that the roots are the complex zeros of a polynomial function, then substitute the given roots in the function and the value of function should come out to be zero.

The value of the imaginary number iota (i) is $\begin{array}{l}i=\sqrt{-1}\\ i^2=-1\end{array}$ .

Proof:

Consider the function, $f\left(x\right)=x^2+6$ .

Substitute $x=-\sqrt{6}i$ in the function $f\left(x\right)=x^2+6$ as,

  $\begin{array}{c}f\left(x\right)=x^2+6\\ f\left(-\sqrt{6}i\right)={\left(-\sqrt{6}i\right)}^2+6\\ =6\left(-1\right)+6\\ =0\end{array}$

Now, substitute $x=\sqrt{6}i$ in the function $f\left(x\right)=x^2+6$ as,

  $\begin{array}{c}f\left(x\right)=x^2+6\\ f\left(\sqrt{6}i\right)={\left(\sqrt{6}i\right)}^2+6\\ =6\left(-1\right)+6\\ =0\end{array}$

Since, both the roots give value of the function as zero; thus, it is verified that given roots $-\sqrt{6}i,\sqrt{6}i$ are the complex zeros of the polynomial function $f\left(x\right)=x^2+6$ .