Chapter 2.5, Problem 15E

To determine

Find the zeros of the function $f(x)=x^4+4x^2+4$ and find the relationship between the number of real zeros and the number of x-intercepts of the graph.

  

Answer to Problem 15E

  $-\sqrt{2}i,-\sqrt{2}i,\sqrt{2}i,\sqrt{2}i$

Explanation of Solution

Given:

Function: $f(x)=x^4+4x^2+4$

  

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=x^4+4x^2+4\\ \Rightarrow f(x)=x^4+4+4x^2\\ \Rightarrow f(x)={(}^{x^2}+2^2+2\times 2\times x^2\\ \Rightarrow f(x)={(}^{x^2}\\ \Rightarrow f(x)=(x^2+2)(x^2+2)\end{array}$

To find the roots of given function, put $f(x)=0,$

  $\begin{array}{l}\Rightarrow (x^2+2)(x^2+2)=0\\ \\ \begin{array}{cc}\begin{array}{l}\Rightarrow x^2+2=0\\ \Rightarrow x^2=-2\\ \Rightarrow x=\pm \sqrt{-2}\\ \Rightarrow x=\pm \sqrt{2}i\\ \Rightarrow x=\sqrt{2}i,-\sqrt{2}i\end{array}& \begin{array}{l}\Rightarrow x^2+2=0\\ \Rightarrow x^2=-2\\ \Rightarrow x=\pm \sqrt{-2}\\ \Rightarrow x=\pm \sqrt{2}i\\ \Rightarrow x=\sqrt{2}i,-\sqrt{2}i\end{array}\end{array}\\ \\ \Rightarrow x=-\sqrt{2}i,-\sqrt{2}i,\sqrt{2}i,\sqrt{2}i\end{array}$

Interpretation from given graph:

The number of real zeros of the function is equal to 0.

Also, the number of x-intercepts is equal to 0.

So, the number of real zeros and number of x-intercepts are equal to each other.

Conclusion:

Therefore, the zeros of given function are $x=-\sqrt{2}i,-\sqrt{2}i,\sqrt{2}i\text{and}\sqrt{2}i.$