Question

Which of the following polynomials does not have real roots?

A. X^2 -17 -7
B. X^2 +4x +8
C. -x^2 +4x +6
D. 2x^2 -13

Answer 1

You get a real root whenever the graph intercepts the x-axis. So you can graph each one to find out.

Answer 2

x1+x2= -b/a

If one of the root is a complex it means the sum would be a complex number,so we have for

a) a=1
b= -17 => x1+x2= -(-17)/1= 17 -it has real roots

b)a=1
b=4 => x1+x2= -4/1= -4 -it has real roots

c)a=-1
b=4 => x1+x2= -4/-1= 4 -it has real roots

d)a=2
b= 0 => x1+x2= -0/2 which doesn't make sense so the answer is variant d)

Answer 3

Another way is to use the determinant (b^2 - 4ac), from the quadratic formula. If it comes out negative there are no real roots.
This works because in the quadratic formula \[x = \pm \frac{ \sqrt{b^2-4ac} }{ 2a }\], if the square root is negative it will have imaginary roots, because you can't have negative square root

Answer 4

Thank you @mariamikayla

Answer 5

You're welcome :D