Question

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

Answer 1

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.

Answer 2

Please see below.

Explanation:

Here,

`0=-3x^2+x-4`

`i.e. 3x^2-x+4=0to`quadratic 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` !!!!!!!