Question 427879


Solving by the Quadratic Formula


Since this formula is somewhat long and complicated, it is best to evaluate it in two smaller pieces by first evaluating the thing inside the radical,

{{{b^2-4ac}}}

and then put the result into the formula 

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

The quantity
b^2-4ac

even has a name. 
It is called the discriminant. And there is another advantage to computing it first. Since it is what is in the radical, it can't be negative if there are going to be solutions to the equation, because you can't take a square root of a negative number, (unless you use the imaginary numbers, and we're not yet ready for them here) so if the discriminant comes out negative, then you don't have to do any more work, and all you have to do is write "no solution" on your paper and you are done. Sometimes you can determine this quite quickly by estimating, particularly if a and c are very large and b is small. 

Example 
Problem: Solve the equation. 

{{{6x^2 -2x+27=0}}}

Solution: 

here we have no solution because {{{b}}} is {{{-2}}}, which is very small in comparison to {{{a=6}}} and {{{c=27}}}, so you don't {{{even}}}{{{ have}}} to compute the discriminant to {{{see }}}that it is going to be {{{negative}}}.