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yanalaym [24]
3 years ago
6

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, 49

i D) x = i + 7, i – 7
Mathematics
1 answer:
malfutka [58]3 years ago
6 0

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 \sqrt{-49}

x=\pm \sqrt{-1}\sqrt{49}

x=\pm 7i                [\because \sqrt{-1}=i]

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

Therefore, the correct option is B.

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