The Quadratic Formula

We know how to solve equations like x^2-7x+12=0. We do this by factoring. If we factor we will get

    \[(x-3)(x-4)=0\]

  so x=3,4

Not all trinomials are factorable. Sometimes we also just can’t figure out how to factor a trinomial. What if we couldn’t figure out how to factor x^2+3x-10=0? We can’t find 2 numbers that multiply to -10 and add to +3. So how are we possibly going to solve this equation? We will use the Quadratic Formula

This is how it works. The generic case looks like ax^2+bx+c=0. So in the case of x^2+3x-10=0

a=1
b=3

c=-10

We take these numbers and plug them into the following formula 

    \[x=\frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)}\]

The \pm symbol means that you have to do this twice; once as a plus and once as a minus. This is where your two answers come from. Let’s plug our equation’s numbers into the quadratic formula to find the answers.

    \[\frac{-3 \pm \sqrt{3^2-4(1)(-10)}}{2(1)}\]

Simplify the insides of the radical we get

    \[\frac{-3 \pm \sqrt{9- -40}}{2}=\frac{-3 \pm \sqrt{49}}{2}= \frac{-3 \pm 7}{2} = \frac{-10}{2}, \frac{4}{2} = -5, 2\]

This can work on any quadratic equation, but the answers may not be so nice.

5 Step Process for Quadratic Formula

 

(Only if you’re having trouble.)

    \[x=\frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)}\]

\textbf{Example:} Solve the equation 2x^2-x-6=0

\textbf{Step 1:} Determine a, b, and c

a=2

b=-1

c=-6

Step 2: Plug into the quadratic formula

    \[\frac{1 \pm \sqrt{(-1)^{2}-4(2)(-6)}}{2(2)}\]

Step 3: Simplify the radical

    \[\sqrt{(-1)^{2}-4(2)(-6)}=\sqrt{49}=\mathbf{7}\]

Step 4: Split into 2 problems — plus & minus

    \[\frac{1 + 7}{2(2)}, \frac{1-7}{2(2)}\]

Step 5: Obtain your two answers

    \[\frac{8}{4}=2 , \frac{-6}{4}=-\frac{3}{2}\]

    \[\mathbf{x=2, -\frac{3}{2}}\]

 

Proof Coming Soon