Example: Solve the following equation for x

0 = x^2 +9x + 20

We can solve for x using factoring if possible or the quadratic equation if we cannot factor. The quadratic formula will always solve quadratic equations for x.

This equation will factor (but note that many quadratic equations will not factor):

(x + 4)(x + 5) = 0

Therefore, we can solve for both values of x

x + 4 = 0

x = -4

AND

x + 5 = 0

x = -5

Solution {-4,-5}

We can also use the quadratic formula and will get the same result:

(-b +/- sqrt(b^2 – 4ac)) / [2a]

For x^2 +9x + 20

a = 1, b = 9, c = 20

(-b +/- sqrt(b^2 – 4ac) / [2a]

(-9 +/- sqrt(9^2 – 4(1)(20)) / [2(1)]

(-9 +/- sqrt (81 – 80) )/ [2]

(-9 +/- sqrt (1) )/ [2]

(-9 +/- 1 )/ [2]

There are two answers:

(-9 -1)/2 = -10/2 = -5

AND

(-9 +1)/2 = -8/2 = -4

This is the same solution we got using factoring and is {-4,-5}

To check your answers:

Check that both solutions for x are correct  – plug each back into the original equation.

Checking on -5

0 = x^2 +9x + 20

0 = (-5)^2 +9(-5) + 20

0 = 25 – 45 + 20

0 = -20 + 20

0 = 0 YES

So -5 is a solution

 

Checking on -4

0 = x^2 +9x + 20

0 = (-4)^2 +9(-4) + 20

0 = 16 – 36 + 20

0 = -20 + 20

0 = 0 YES

So -4 is a solution