Lacey's question at Yahoo Answers regarding a quadratic equation

[MarkFL]

Here is the question:

Solve the quadratic equation?

Consider the following quadratic equation -3x^2-x-5=0

Use the discriminant b^2-4ac to determine the number of solutions of the quadratic equation

Solve using the quadratic equation usinf the formula y= (-b +/- √b^2-4ac) / 2a

Here is a link to the question:

Solve the quadratic equation? - Yahoo! Answers

I have posted a link there so the OP can find my response.

 

[MarkFL]

Hello Lacey,

The first thing I would do is multiply through by -1, just to get rid of all those negative signs, as this will not change the value of the discriminant and roots:

\(\displaystyle 3x^2+x+5=0\)

Now, we identify:

\(\displaystyle a=3,\,b=1,\,c=5\)

and the discriminant is:

\(\displaystyle \Delta=b^2-4ac=(1)^2-4(3)(5)=1-60=-59<0\)

Since the discriminant is negative, we know we will have two, complex conjugate roots.

Using the quadratic formula, we find that these roots are:

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{-59}}{2(3)}=\frac{-1\pm i\sqrt{59}}{6}\)

To Lacey and any other guests viewing this topic, I invite and encourage you to post other such questions in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.