Quadratic Formula

Some quadratic equations cannot be factored easily. For example, in the equation $x^2+3x+1=0$, we need to find two factors of $1$ that add to $3$. But the only integer factors of $1$ are $1$ and $1$, and they definitely don’t add to $3$! To solve these hard-to-factor quadratic equations, it’s easiest to use the quadratic formula given below, which tells us explicitly how to compute the solutions of a quadratic equation $ax^2+bx+c=0$.

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


Using the quadratic formula, we can compute the solutions to the equation $x^2+3x+1=0$.

$\begin{align*} \begin{matrix} \text{Substitute} a=1 \text{, }b=3\text{, and} \\ c=1 \text{ in quadratic formula} \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} x = \frac{-3 \pm \sqrt{ (3)^2 - 4(1)(1)} }{2(1)} \\ \text{Simplify} \hspace{.5cm} &\Bigg| \hspace{.5cm} x = \frac{-3 \pm \sqrt{5} }{2} \\ \begin{matrix} \text{Separate the } \pm \text{ into two} \\ \text{solutions (optional)} \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} x = \frac{-3 + \sqrt{5} }{2} \text{ or } x = \frac{-3 - \sqrt{5} }{2} \end{align*}$


These solutions look weird, but they’re correct.

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To gain some faith in the quadratic formula, we can also rearrange it back into the original equation to see that it must have the same solutions as the original equation:

$\begin{align*} x &= \frac{-b \pm \sqrt{b^2-4ac} }{2a} \\ 2ax &= -b \pm \sqrt{ b^2 - 4ac} \\ 2ax+b &= \pm \sqrt{ b^2 - 4ac} \\ (2ax+b)^2 &= \left( \pm \sqrt{ b^2 - 4ac} \right)^2 \\ 4a^2x^2 + 4abx + b^2 &= b^2 - 4ac \\ 4a^2x^2 + 4abx + 4ac &= 0 \\ 4a(ax^2 + bx + c ) &= 0 \\ ax^2 + bx + c &= 0 \end{align*}$


Using the quadratic equation, we can see that some quadratic equations have 2 solutions (as usual), but other quadratic equations can have just 1 solution, or no solutions at all.

For example, using the quadratic equation to solve $x^2+2x+1$, we find a single solution because the $\pm \sqrt{b^2 - 4ac}$ part comes out to $\pm 0$.

$x = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(1)}}{2(1)} = \dfrac{-2 \pm 0}{2} = -1$


Similarly, using the quadratic equation to solve $x^2+2x+2$, we find no solutions because the $\pm \sqrt{b^2-4ac}$ part comes out to $\pm \sqrt{ \text{negative number} }$, and we can’t take the square root of a negative number. (We’ll ignore imaginary solutions and consider only real solutions for now.)

$x = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \dfrac{-2 \pm \sqrt{-4} }{2} = \text{no solution}$


To see how many solutions a quadratic equation has, we need only consider the $b^2-4ac$ part of the quadratic formula, which is called the discriminant. If the discriminant is positive, then we have two solutions. If it is $0$, then we have one solution. If it is negative, then we have no solution.

We can also use the quadratic formula to understand the trick for factoring when $a$ is not equal to $1$ – which was to multiply $c$ by $a$, replace $a$ with $1$, factor the result, divide each constant in each factor by the original $a$, and move denominators onto our variables. From the quadratic formula, we know that the solutions of $ax^2+bx+c=0$ are given by $x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$. When we multiply $c$ by $a$ and replace $a$ with $1$, we have the equation $x^2+bx+ac=0$, which has solutions $x= \frac{-b \pm \sqrt{b^2-4ac}}{2}$. This means that if $x$ is a solution of $x^2+bx+ac=0$, then $ax$ is a solution of $ax^2+bx+c$. Thus, if $x^2+bx+ac$ factors into $(x+m)(x+n)$, then $ax^2+bx+c$ factors into $a \left( x+\frac{m}{a} \right) \left( x+\frac{n}{a} \right)$.


Practice Problems


Use the quadratic formula to solve the following quadratic equations. (You can view the solution by clicking on the problem.)

$1) \hspace{.5cm} x^2+3x-7=0$
Solution:
$x = \frac{-3 \pm \sqrt{37}}{2}$

$2) \hspace{.5cm} 4x^2-12x+9=0$
Solution:
$x = \frac{3}{2}$

$3) \hspace{.5cm} -2x^2+4x+6=0$
Solution:
$x = -1 \text{ or } x=3$

$4) \hspace{.5cm} 3x^2-x+5=0$
Solution:
$\text{no solution}$

$5) \hspace{.5cm} -25x^2-20x=4$
Solution:
$x = -\frac{2}{5}$

$6) \hspace{.5cm} 3x^2=2x+3$
Solution:
$x = \frac{1 \pm \sqrt{10}}{3}$

$7) \hspace{.5cm} 42x=9x^2+49$
Solution:
$x = \frac{7}{3}$

$8) \hspace{.5cm} 8x^2+5=3x$
Solution:
$\text{no solution}$

$9) \hspace{.5cm} 1+3x=5x^2$
Solution:
$x = \frac{3 \pm \sqrt{29}}{10}$

$10) \hspace{.5cm} 154x=121x^2+49$
Solution:
$x = \frac{7}{11}$

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