Question

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

Answer 1

B

Explanation:

To find the zeros equate f(x) to zero , that is

x² + 49 = 0 ( subtract 49 from both sides )

x² = - 49 ( take the square root of both sides )

x = ±
`√(-49)` = ± 7i

Solutions are x = - 7i, x = 7i

Answer 2

B

Explanation:

f(x) = x²+49 , rewrite the equation

x² +7² =0 , to find the roots make f(x) =0 and subtract 7² from both sides

x² = -7², square root both sides

√x² =√-7², consider √-1 =i

x= ±7i

the roots are -7i and 7i