Quadratic formula is used in order to find the roots (solution) of a second degree (quadratic) function. By solution we mean x values, where the function crosses the x-axis, which is the same as x-intercept. In order to find the x-intercept, you should set the y (which is your function) equal to zero. That is why we have:
ax^2 + bx + c = 0
Quadratic functions can have 0,1, or 2 solutions. This depends on the discriminant. Before getting to that, remember that the formula to find the solution of a second degree (quadratic) equation is given by:
x = (-b +/- sqrt (b^2-4ac)/ 2a where sqrt() means square root.
In the above formula, we call the term b^2-4ac the discriminant. It is the value of the discriminant that determines whether there is 0, 1, or 2 solutions. This is how we can tell: If the value discriminant comes out to be negative, it implies that there are no solutions (real solutions of course, but you would have imaginary solution). If the value of discriminant comes out to be 0, that means you will have only one real solution. And if the value is positive, you will have 2 real solutions.
Now, let's look at the simplest case of a quadratic function: y= x^2
You probably know how the graph of y=x^2 should look like. When x=0, y= 0^2 =0 so when x=0, y=0. There are no other values of x that make the function (y) equal to 0. Therefore there is only one solution which is x=0.
Now let's use the formula to confirm:
y= x^2
comparing it to the standard equation y= ax^2+bx+c, we have: a=1 , b=0, c=0 (because we do not have bx+c term in our equation, so both b and c would be 0, and the coefficient of x^2 is 1 so therefore a=1).
the discriminant is: b^2 - 4ac = 0^2 - 4(1*0) = 0-0=0
discriminant is 0, so we only have one solution!
using the quadratic formula, x = (-b +/- sqrt (b^2-4ac)/ 2a
It simplifies to x= -b/2a (we got rid of discriminant because it turned out to be 0)
so x= -0/(2*1) =0
so x=0 is the solution, as we expected.
Please let me know if you could follow this well, and if so I can give you other examples that are slightly more complicated than this. However, the logic and procedure is the same in all of them, so you can even do the harder ones just by following this method and looking at some examples online.
Good luck!
Kaveh
