Solving Quadratic Equations Using the Quadratic Formula

Any quadratic equation can be simplified so that it is of the form:

$ax^2+bx+c=0$

where a,b, and c are real numbers and a is not zero. Once the equation is written in this form, the solution(s) can be found using the quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

When the numbers are plugged into this formula, and the resulting answer simplified, it may be found that there are no real solutions (in other words there are complex solutions) , one real solution, or two real solutions. Often, you want an exact answer – not a decimal answer, so there is a bit of work to be done in simplifying.

Example Solve $x^2-x-6=0$ .

Solution

In this example, $a=1$ , $b= -1$ , and $c=-6$ . Why are $b$ and $c$ negative? The formula is based off the form $ax^2+bx+c=0$ , so if you come across anything that is negative, keep the sign when you use the formula!

$x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}$

$=\dfrac{1\pm\sqrt{1+24}}{2}$

$=\dfrac{1\pm\sqrt{25}}{2}$ .

At this stage, the plus or minus symbol ( $\pm$ ) tells you that there are actually two different solutions:

$\dfrac{1+\sqrt{25}}{2}=\dfrac{1+5}{2}=\dfrac{6}{2}=3$ and

$\dfrac{1- \sqrt{25}}{2}=\dfrac{1-5}{2}=\dfrac{-4}{2}=-2$ .

Therefore the final answer would be $x=3,-2$ .

The example above would be considered one of the “easier” problems with the quadratic formula since the square root was easy to find and the solutions turned out to be whole numbers. This will most often NOT happen and when it doesn’t , you will need to remember how to simplify radical terms as in the next example.

Example Solve $2x^2+2x-7=0$

Solution:

In this case, $a=2$ , $b=2$ , and $c=-7$ . Therefore

$x=\dfrac{-2\pm\sqrt{(-2)^2-4(2)(-7)}}{2(2)}$

$=\dfrac{-2\pm\sqrt{4+56}}{4}$

$=\dfrac{-2\pm\sqrt{60}}{4}$

$=\dfrac{-2\pm 2\sqrt{15}}{4}$

Notice that $2$ is a FACTOR of both the numerator and denominator, so it can be cancelled.

$=\dfrac{-1\pm\sqrt{15}}{2}$ .

This answer can not be simplifed anymore without using decimals. Therefore the final answer is: $x=\dfrac{-1+\sqrt{15}}{2}$ , $\dfrac{-1-\sqrt{15}}{2}$ .