Using the Discriminant to Predict the Number of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant.

Discriminant

In the Quadratic Formula \(x=\frac{\text{−}b±\sqrt{{b}^{2}-4ac}}{2a}\), the quantity \({b}^{2}-4ac\) is called the discriminant.

Let’s look at the discriminant of the equations in examples from the previous lesson, and the number of solutions to those quadratic equations.

	Quadratic Equation (in standard form)	Discriminant \({b}^{2}-4ac\)	Sign of the Discriminant	Number of real solutions
Example 1	\(2{x}^{2}+9x-5=0\)	\({9}^{2}-4·2(-5)=121\)	+	2
Example 2	\(4{x}^{2}-20x+25=0\)	\({(-20)}^{2}-4·4·25=0\)	0	1
Example 3	\(3{p}^{2}+2p+9=0\)	\({2}^{2}-4·3·9=-104\)	−	0

When the discriminant is positive\((x=\frac{\text{−}b±\sqrt{+}}{2a})\) the quadratic equation has two solutions.

When the discriminant is zero\((x=\frac{\text{−}b±\sqrt{0}}{2a})\) the quadratic equation has one solution.

When the discriminant is negative\((x=\frac{\text{−}b±\sqrt{-}}{2a})\) the quadratic equation has no real solutions.

Use the discriminant, \({b}^{2}-4ac\), to determine the number of solutions of a Quadratic Equation.

For a quadratic equation of the form \(a{x}^{2}+bx+c=0\), \(a\ne 0\),

if \({b}^{2}-4ac>0\), the equation has two solutions.
if \({b}^{2}-4ac=0\), the equation has one solution.
if \({b}^{2}-4ac<0\), the equation has no real solutions.

Example

Determine the number of solutions to each quadratic equation:

\(2{v}^{2}-3v+6=0\)
\(3{x}^{2}+7x-9=0\)
\(5{n}^{2}+n+4=0\)
\(9{y}^{2}-6y+1=0\)

Solution

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

\(\begin{array}{cccc}& & & \hfill \phantom{\rule{5em}{0ex}}2{v}^{2}-3v+6=0\\ \text{The equation is in standard form, identify}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill \phantom{\rule{5em}{0ex}}a=2,b=-3,c=6\\ \text{Write the discriminant.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}{b}^{2}-4ac\phantom{\rule{1.3em}{0ex}}\\ \text{Substitute in the values of}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill \phantom{\rule{5em}{0ex}}{(3)}^{2}-4·2·6\\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\begin{array}{c}\hfill 9-48\phantom{\rule{1.7em}{0ex}}\\ \hfill -39\phantom{\rule{2.1em}{0ex}}\end{array}\\ \text{Because the discriminant is negative, there are no real}\hfill & & & \\ \text{solutions to the equation.}\hfill & & & \end{array}\)
\(\begin{array}{cccc}& & & \hfill \phantom{\rule{6.2em}{0ex}}3{x}^{2}+7x-9=0\\ \text{The equation is in standard form, identify}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill \phantom{\rule{6.2em}{0ex}}a=3,b=7,c=-9\\ \text{Write the discriminant.}\hfill & & & \hfill \phantom{\rule{6.2em}{0ex}}{b}^{2}-4ac\phantom{\rule{2.6em}{0ex}}\\ \text{Substitute in the values of}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill \phantom{\rule{6.2em}{0ex}}{(7)}^{2}-4·3·(-9)\\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{6.2em}{0ex}}\begin{array}{c}\hfill 49+108\phantom{\rule{2.6em}{0ex}}\\ \hfill 157\phantom{\rule{3.4em}{0ex}}\end{array}\\ \text{Because the discriminant is positive, there are two}\hfill & & & \\ \text{solutions to the equation.}\hfill & & & \end{array}\)
\(\begin{array}{cccc}& & & \hfill \phantom{\rule{5em}{0ex}}5{n}^{2}+n+4=0\\ \text{The equation is in standard form, identify}\phantom{\rule{0.2em}{0ex}}a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c.\hfill & & & \hfill \phantom{\rule{5em}{0ex}}a=5,b=1,c=4\\ \text{Write the discriminant.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}{b}^{2}-4ac\phantom{\rule{1.2em}{0ex}}\\ \text{Substitute in the values of}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill \phantom{\rule{5em}{0ex}}{(1)}^{2}-4·5·4\\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\begin{array}{c}\hfill 1-80\phantom{\rule{1.7em}{0ex}}\\ \hfill -79\phantom{\rule{2.1em}{0ex}}\end{array}\\ \text{Because the discriminant is negative, there are no real}\hfill & & & \\ \text{solutions to the equation.}\hfill & & & \end{array}\)
\(\begin{array}{cccc}& & & \hfill 9{y}^{2}-6y+1=0\\ \text{The equation is in standard form, identify}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill a=9,b=-6,c=1\\ \text{Write the discriminant.}\hfill & & & \hfill {b}^{2}-4ac\phantom{\rule{1.2em}{0ex}}\\ \text{Substitute in the values of}\phantom{\rule{0.2em}{0ex}}a,b,c.\hfill & & & \hfill {(-6)}^{2}-4·9·1\\ \text{Simplify.}\hfill & & & \hfill \begin{array}{c}\hfill 36-36\phantom{\rule{1.6em}{0ex}}\\ \hfill 0\phantom{\rule{2.7em}{0ex}}\end{array}\\ \text{Because the discriminant is 0, there is one solution to the equation.}\hfill & & & \end{array}\)

Using the Discriminant to Predict the Number of Solutions of a Quadratic Equation