How do you find the complex roots of $6 c^{3} - 3 c^{2} - 45 c = 0$ $6c^3-3c^2-45c=0$ ?

Question

How do you find the complex roots of $6 c^{3} - 3 c^{2} - 45 c = 0$ $6c^3-3c^2-45c=0$ ?

1 Answer

Impact of this question

2207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-27T05:37:03

Remove a common factor of $3 c$ $3c$ from each term.

$6 c^{3} - 3 c^{2} - 45 c = 0$ $6c^3-3c^2-45c=0$

$3 c (2 c^{2} - c - 9) = 0$ $3c(2c^2-c-9)=0$

Checking the discriminant of the resulting quadratic, $Delta=b^2-4ac=(-1)^2-4(2)(-9)=73$ we see that the quadratic will have two Real roots for a total of three Real roots (and no Complex roots).

The first zero comes from $3c=0$ , or $c=0$ .

The second two, through the quadratic formula for $2c^2-c-9=0$ are:

$c=(1+-sqrt(1-4(2)(-9)))/4=(1+-sqrt73)/4$

We see that our three zeros are $c=0,(1+-sqrt73)/4$ and that these are all three Real zeros.

Science

Math

Humanities

... and beyond

How do you find the complex roots of $6 c^{3} - 3 c^{2} - 45 c = 0$ $6c^3-3c^2-45c=0$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the complex roots of 6c^3-3c^2-45c=06c3−3c2−45c=06c^3-3c^2-45c=0?

1 Answer

Related questions

Impact of this question

How do you find the complex roots of $6 c^{3} - 3 c^{2} - 45 c = 0$ $6c^3-3c^2-45c=0$ ?