Given the polynomial function [tex]\( f(x) \)[/tex] with rational coefficients, we are informed that two of its roots are 3 and [tex]\( \sqrt{7} \)[/tex].
When dealing with polynomials that have rational coefficients, a crucial property is that any irrational root must have its conjugate also as a root. The same concept applies to complex roots and their conjugates as follows:
1. If [tex]\( a + b\sqrt{c} \)[/tex] is a root, then [tex]\( a - b\sqrt{c} \)[/tex] must also be a root.
2. If [tex]\( a + bi \)[/tex] (complex root) is a root, then [tex]\( a - bi \)[/tex] must also be a root.
For the given polynomial:
- One of the roots is [tex]\( 3 \)[/tex]. Since 3 is rational, there is no additional root derived directly from it.
- Another root given is [tex]\( \sqrt{7} \)[/tex]. This is an irrational root.
Given that [tex]\( \sqrt{7} \)[/tex] is an irrational number, the property dictates that the conjugate [tex]\( -\sqrt{7} \)[/tex] must also be a root of the polynomial to ensure the coefficients remain rational.
Therefore, examining the given options for the roots that must also be present in the polynomial [tex]\( f(x) \)[/tex]:
- [tex]\( -\sqrt{7} \)[/tex]: This is the rational conjugate of [tex]\( \sqrt{7} \)[/tex], and hence it must indeed be a root.
- [tex]\( i\sqrt{7} \)[/tex]: This would introduce imaginary coefficients, conflicting with the requirement of rational coefficients.
- [tex]\( -3 \)[/tex]: There is no requirement from the given roots that [tex]\( -3 \)[/tex] must be a root.
- [tex]\( 3i \)[/tex]: Similarly, this would introduce complex coefficients, which are not allowed since the polynomial has rational coefficients.
Thus, the correct root that must also be part of [tex]\( f(x) \)[/tex] in addition to the given roots is:
[tex]\[ -\sqrt{7} \][/tex]
Therefore, the answer is:
[tex]\[ -\sqrt{7} \][/tex]
