How many real number solutions are there to the equation 0=-3x^2+x-4?

2 Answers
Jul 2, 2018

No real solutions it has.

Explanation:

0 = -3x^2 + x - 4

Doing the first step of quadratic formula, i.e. finding the discriminant gives us the answer:

"discriminant", d = b^2 - 4ac
if d < 0, no real solution,
if d = 0, one solution
and if d > 0, the quadratic equation has two solutions.

In this case, d = 1^2 - 4(-3)(-4)
d = -47
It has no real roots.

Jul 2, 2018

Please see below.

Explanation:

Here,

0=-3x^2+x-4

i.e. 3x^2-x+4=0toquadratic equation

Comparing with ax^2+bx+c=0

a=3 ,b=-1 and c=4

=>Discriminant triangle=b^2-4ac=(-1)^2-4(3)(4)

=>color(red)(triangle=1-48=-47 < 0=>x !inRR

or triangle < 0=>x in CC and complex conjugate roots.

Hence , the equation has no real root , but two complex roots.

i.e. two solutions in CC !!!!!!!