Some quadratic equations cannot be easily solved by either factoring or by the square root property, as discussed in prior lessons. However, using a method known as completing the square, one can find the solution to any quadratic equation.
In fact, if given a quadratic equation of the form
$$ax^2+bx+c=0$$
where \(a,b\) and \(c\) are any real numbers (except \(a=0\)), then the solution to this equation is the pair of \(x\)-values
$$ x=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ and $$x=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
or, more succinctly in one single formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
The above is known as the Quadratic Formula.
What you get out of this formula are (potentially) two \(x\)-values that, when substituted into the left hand side of the above equation, gives you 0. These solutions to the quadratic equation are also known as roots for the quadratic expression \(ax^2+bx+c\).
We care because in many real-world applications of quadratic equations, you are not given nice and easy-to-factor quadratic expressions. More often, you are given equations like \(0.421x^2+12.313x=0.3638\) and are asked to find what \(x\)-values make that equation true (i.e. make the left-hand-side equal to zero). Good luck factoring that loveliness! Instead, just plug the coefficients into the Quadratic Formula. Namely, let
and plug these coefficients into the quadratic formula to get
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(12.313)\pm\sqrt{(12.313)^2-4(0.421)(0.3638)}}{2(0.421)}$$
Noting that the \(\pm\) gives us two different \(x\)-values, namely
$$\frac{-(12.313)+\sqrt{(12.313)^2-4(0.421)(0.3638)}}{2(0.421)}=-0.0295759$$
and
$$\frac{-(12.313)-\sqrt{(12.313)^2-4(0.421)(0.3638)}}{2(0.421)}=-29.2175$$
Before using the quadratic formula to solve your quadratic equation, you must first make sure that everything that is NOT zero is all on one side of your equation, and that there is a zero on the other side. THEN you will be able to find the correct coefficients \(a,b\) and \(c\) to use in the quadratic formula.