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Discriminant

Determine the number of real roots using the discriminant

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Using the Discriminant

The profit on your school fundraiser is represented by the quadratic expression \begin{align*}-5p^2 + 400p - 8000\end{align*}5p2+400p8000, where p is your price point. How many real break-even points will you have?

Discriminant

Remember, the Quadratic Formula is \begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}x=b±b24ac2a. The expression under the radical, \begin{align*}b^2-4ac\end{align*}b24ac, is called the discriminant. You can use the discriminant to determine the number and type of solutions an equation has.

Solving Equations with Different Types of Solutions

Step 1: Solve \begin{align*}x^2-8x-20=0\end{align*}x28x20=0 using the Quadratic Formula. What is the value of the discriminant?

\begin{align*}x &=\frac{8 \pm \sqrt{{\color{red}144}}}{2}\\ &=\frac{8 \pm 12}{2} \rightarrow 10, -2\end{align*}x=8±1442=8±12210,2

Step 2: Solve \begin{align*}x^2-8x+6=0\end{align*}x28x+6=0 using the Quadratic Formula. What is the value of the discriminant?

\begin{align*}x &=\frac{8 \pm \sqrt{{\color{red}0}}}{2}\\ &=\frac{8 \pm 0}{2} \rightarrow 4\end{align*}x=8±02=8±024

Step 3: Solve \begin{align*}x^2-8x+20=0\end{align*}x28x+20=0 using the Quadratic Formula. What is the value of the discriminant?

\begin{align*}x &=\frac{8 \pm \sqrt{{\color{red}-16}}}{2}\\ &=\frac{8 \pm 4i}{2} \rightarrow 4 \pm 2i\end{align*}x=8±162=8±4i24±2i

Step 4: Look at the values of the discriminants from Steps 1-3. How do they differ? How does that affect the final answer?

From the steps above, we can conclude:

  • If \begin{align*}b^2-4ac > 0\end{align*}b24ac>0, then the equation has two real solutions.
  • If \begin{align*}b^2-4ac = 0\end{align*}b24ac=0, then the equation has one real solution; a double root.
  • If \begin{align*}b^2-4ac < 0\end{align*}b24ac<0, then the equation has two imaginary solutions.

   

 

Let's determine the type of solutions for the following equations.

  1. \begin{align*}4x^2-5x+17=0\end{align*}4x25x+17=0

Find the discriminant.

\begin{align*}b^2-4ac &=(-5)^2-4(4)(17)\\ &=25-272\end{align*}b24ac=(5)24(4)(17)=25272

At this point, we know the answer is going to be negative, so there is no need to continue (unless we were solving the problem). This equation has two imaginary solutions.

Use the Quadratic Formula for the previous problem. 

\begin{align*}x = \frac{5 \pm \sqrt{25-272}}{8}=\frac{5 \pm \sqrt{-247}}{8}= \frac{5}{8} \pm \frac{\sqrt{247}}{8}i\end{align*}x=5±252728=5±2478=58±2478i

  1. \begin{align*}3x^2-5x-12=0\end{align*}3x25x12=0

Use the discriminant. a = 3, b = -5, and c = -12

\begin{align*} \sqrt{(-5)^2 - 4(3)(-12)}=\sqrt{25+144}=\sqrt{169}=13\end{align*}(5)24(3)(12)=25+144=169=13

This quadratic has two real solutions.

Examples

Example 1

Earlier, you were asked to find the number of break-even points will you have. 

Set the expression \begin{align*}-5p^2 + 400p - 8000\end{align*}5p2+400p8000 equal to zero and then find the discriminant.

\begin{align*}-5p^2 + 400p - 8000 = 0\end{align*}5p2+400p8000=0

\begin{align*}b^2-4ac &=(400)^2-4(-5)(-8000)\\ &=160000-160000 = 0\end{align*}b24ac=(400)24(5)(8000)=160000160000=0

At this point, we know the answer is zero, so the equation has one real solution. Therefore, there is one real break-even point.

Example 2

Use the discriminant to determine the type of solutions \begin{align*}-3x^2-8x+16=0\end{align*}3x28x+16=0 has.

\begin{align*}b^2-4ac &=(-8)^2-4(-3)(16)\\ &=64+192\\ &=256\end{align*}b24ac=(8)24(3)(16)=64+192=256

This equation has two real solutions.

Example 3

Use the discriminant to determine the type of solutions \begin{align*}25x^2-80x+64=0\end{align*}25x280x+64=0 has.


\begin{align*}b^2-4ac &=(-80)^2-4(25)(64)\\ &=6400-6400\\ &=0\end{align*}b24ac=(80)24(25)(64)=64006400=0

This equation has one real solution.

Example 4

Solve the equation from Example 2. 

\begin{align*}x = \frac{8 \pm \sqrt{256}}{-6}=\frac{8 \pm 16}{-6}=-4, \frac{4}{3}\end{align*}x=8±2566=8±166=4,43

Review

Determine the number and type of solutions each equation has.

  1. \begin{align*}x^2-12x+36=0\end{align*}x212x+36=0
  2. \begin{align*}5x^2-9=0\end{align*}5x29=0
  3. \begin{align*}2x^2+6x+15=0\end{align*}2x2+6x+15=0
  4. \begin{align*}-6x^2+8x+21=0\end{align*}6x2+8x+21=0
  5. \begin{align*}x^2+15x+26=0\end{align*}x2+15x+26=0
  6. \begin{align*}4x^2+x+1=0\end{align*}4x2+x+1=0

Solve the following equations using the Quadratic Formula.

  1. \begin{align*}x^2-17x-60=0\end{align*}x217x60=0
  2. \begin{align*}6x^2-20=0\end{align*}6x220=0
  3. \begin{align*}2x^2+5x+11=0\end{align*}2x2+5x+11=0

Challenge Determine the values for \begin{align*}c\end{align*}c that make the equation have a) two real solutions, b) one real solution, and c) two imaginary solutions.

  1. \begin{align*}x^2+2x+c=0\end{align*}x2+2x+c=0
  2. \begin{align*}x^2-6x+c=0\end{align*}x26x+c=0
  3. \begin{align*}x^2+12x+c=0\end{align*}x2+12x+c=0
  4. What is the discriminant of \begin{align*}x^2+2kx+4=0\end{align*}? Write your answer in terms of \begin{align*}k\end{align*}.
  5. For what values of \begin{align*}k\end{align*} will the equation have two real solutions?
  6. For what values of \begin{align*}k\end{align*} will the equation have one real solution?
  7. For what values of \begin{align*}k\end{align*} will the equation have two imaginary solutions?

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.14. 

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