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

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Answer 1 · 2018-07-02T07:55:36

No real solutions it has.

#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 · 2018-07-02T08:29:06

Please see below.

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# !!!!!!!