• April 13th 2008, 09:31 AM
erinspice
The Quadratic Formula Explained says:

Quote:

So, while factoring may not always be successful, the Quadratic Formula can always find the solution.

I'm trying to use the Quadratic Formula to solve
$1.26x^2-186.56x+34041.94=0$ but my $4ac$ is much bigger than my $b^2$ so everything under the radical evaulates to a negative number. How can I solve this?
• April 13th 2008, 09:34 AM
Jhevon
Quote:

Originally Posted by erinspice
The Quadratic Formula Explained says:

I'm trying to use the Quadratic Formula to solve
$1.26x^2-186.56x+34041.94=0$ but my $4ac$ is much bigger than my $b^2$ so everything under the radical evaulates to a negative number. How can I solve this?

it IS possible for a quadratic to have no real solutions. Look at x^2 + 1 = 0. This is never true for any real number x. Have you learned about complex numbers, and complex solutions? If so, the quadratic formula can still find those
• April 13th 2008, 09:44 AM
erinspice
Quote:

Originally Posted by Jhevon
it IS possible for a quadratic to have no real solutions. Look at x^2 + 1 = 0. This is never true for any real number x. Have you learned about complex numbers, and complex solutions? If so, the quadratic formula can still find those

Yeah, but I need a real solution. My robot won't know what to do with a distance of xi (see this thread)! I must have made an error somewhere, because I know there has to be a real solution to my problem.
• April 13th 2008, 09:47 AM
janvdl
Quote:

Originally Posted by erinspice
The Quadratic Formula Explained says:

I'm trying to use the Quadratic Formula to solve
$1.26x^2-186.56x+34041.94=0$ but my $4ac$ is much bigger than my $b^2$ so everything under the radical evaulates to a negative number. How can I solve this?

If you wish to check your answers, even those which have complex roots, download the MHF QuickSolver. The link to it is in my signature.
• April 14th 2008, 10:47 PM
cigster
Complex zeros
If one sketches the quadratic function, one can see that graph of the function does not at any point intersect x-axis thus there is no real zero. However, for polynom of degree 2 there must neccesarily be two zeros. Because they are not real, they must be complex, which I guess in your case is not neccesary to calculate.