Question

How would i start this?

Answer 1

Well you can either use the quadratic formula or you can solve it by factoring. If you can't find the factors of the quadratic equation, it means it doesn't have any x-intercepts or real solutions. Do you know how to apply any of the methods I mentioned?

Answer 2

not so much

Answer 3

Let's start with the radical part of the quadratic formula, to check whether the equation will have a solution or not. We call this as the discriminant:

\(\sf d=b^2-4ac\) where \(\sf ax^2+bx+c=0\)

if \(\sf d\) = 0 , one real root
if \(\sf d\) < 0 , no real root
if \(\sf d\) > 0 , two real roots

So for your first equation, a= 1 b= 0 and c = 4
if you plug it into the discriminant formula, what will it give you?

Answer 4

1^2 + 0x + 4 = 0

Answer 5

\(\sf x^2 + 4 = 0\) this is your first equation...
we need to find out whether it will have a solution

so let's plug the values of a, b and c in the discriminant formula:
\(\sf d= b^2-4ac= 0^2- 4(1)(4)= -16 \)

so d is -16 , now is -16 less than, greater than or is it equal to zero?

Answer 6

sry its late igtg