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tigry1 [53]
3 years ago
5

If a polynomial function f(x) has roots 3 and square root of 7 what must also be a root of f(x)?

Mathematics
2 answers:
Whitepunk [10]3 years ago
8 0

Answer:

-\sqrt{7}

A is correct.

Step-by-step explanation:

A polynomial function f(x) has root 3\text{ and }\sqrt{7}.

3 is a real number.

\sqrt{7} is an irrational number.

The zeros or root of the function always occurs in conjugate pair.

Conjugate pair: A root has two form one positive and one negative.

e.g:a+\sqrt{b},a-\sqrt{b}

For the given function f(x), \sqrt{7} should be in conjugate pair.

One more possible root would be -\sqrt{7}

Hence, the one root must be negative of root of 7

yuradex [85]3 years ago
6 0
Whenever you have a root of sqrt #, you must have the matching - (or positive).
This is because when you take a square root you get two solutions a positive and a negative.
The answer is LETTER A
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