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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