Answer:
Yes, Juanita is correct.
Explanation:
Given:
The function given is:
![f(x)=x^2+64](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3f6pdj5847kdme0denwnzvbpki304lsaj8.png)
The roots of a function means the values of 'x' for which the function intersects or touches the x-axis. Therefore, the roots are nothing but the x-intercepts of the function.
In order to find the roots of the above function, we should equate it to 0.
So,
![f(x)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rur55zla9ujowurgyoxeoex2g38c6hzs1g.png)
![x^2+64=0\\\\x^2=-64](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gpwuioafakdphp9iarg2pib3oguxkjibb2.png)
We know that, square of any number is always a positive quantity. But here, the square of 'x' is -64 which is a negative number. This is not possible.
So, there is no real solution to the given function. In other words, we can say that the function doesn't intersect the x-axis.
Therefore, Juanita is correct in her argument.