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Find all real and non-real roots of the function ƒ(x) = x2 + 49. Question 1 options: A) x = –7, 7 B) x = –7i, 7i C) x = –49i, 49i D) x = i + 7, i – 7

User Daniel Torres
by
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2 Answers

25 votes
25 votes

Answer:

x = –7i, 7i

Explanation:

User Kvark
by
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12 votes
12 votes

Given:

The function is:


f(x)=x^2+49

To find:

The all the real and non-real roots of the given function.

Solution:

We have,


f(x)=x^2+49

For roots,
f(x)=0,


x^2+49=0


x^2=-49

Taking square root on both sides, we get


x=\pm √(-49)


x=\pm √(-1)√(49)


x=\pm 7i
[\because √(-1)=i]

The roots of the given function are -7i and 7i.

Therefore, the correct option is B.

User Per Mikkelsen
by
3.0k points