# How many complex roots does #6x^4+x^3-5=0# have?

##### 2 Answers

#### Explanation:

So

has, at least 1 Real root (

The remaining factor equation:

- has no (apparent) discontinuities
- approaches
#-oo# as#x# decreases - approaches
#+oo# as#x# increases

and therefore most cross the X-axis at some point

An polynomial equation with highest variable exponent of 4 may have 4 roots (some may disappear if two roots occur at the same value).

We have established that 2 of the roots are Real.

And the question is whether the remaining 2 are Real or Complex.

Taking the derivative of

The discriminant of this (using the quadratic formula) indicates that there are no Real roots (

and since there are no points of inflection,

is a continuously increasing function which (therefore) only cross the X-axis at one point.

Therefore, there must be 2 Complex roots.

#### Explanation:

Factoring won't work very well, but the graph of a function is a great indicator of its types of roots.

Since this is a quartic equation (it has degree

Real roots occur when the graph of the function crosses the

The function's graph is:

graph{6x^4+x^3-5 [-4, 4, -6.33, 6.33]}

Here, we can see the two real roots at

Since there are only two real roots, the other two must be complex roots. Note that this makes mathematical sense since complex roots will always come in pairs in polynomials with real coefficients such as this.