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# 10.6: The Discriminant

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the discriminant of a quadratic equation.
• Interpret the discriminant of a quadratic equation.
• Solve real-world problems using quadratic functions and interpreting the discriminant.

## Introduction

The quadratic equation is \begin{align*}ax^2 + bx + c = 0\end{align*}.

It can be solved using the quadratic formula \begin{align*} x = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}\end{align*}.

The expression inside the square root is called the discriminant, \begin{align*}D = b^2 - 4ac\end{align*}. The discriminant can be used to analyze the types of solutions of quadratic equations without actually solving the equation. Here are some guidelines.

• If \begin{align*}b^2 - 4ac > 0\end{align*}, we obtain two separate real solutions.
• If \begin{align*}b^2 - 4ac < 0\end{align*}, we obtain non-real solutions.
• If \begin{align*}b^2 - 4ac = 0\end{align*}, we obtain one real solution, a double root.

## Find the Discriminant of a Quadratic Equation

To find the discriminant of a quadratic equation, we calculate \begin{align*}D = b^2 - 4ac\end{align*}.

Example 1

Find the discriminant of each quadratic equation. Then tell how many solutions there will be to the quadratic equation without solving.

a) \begin{align*}x^2 - 5x + 3 = 0\end{align*}

b) \begin{align*}4x^2 - 4x + 1 = 0\end{align*}

c) \begin{align*}-2x^2 + x = 4\end{align*}

Solution:

a) Substitute \begin{align*}a = 1, b = -5\end{align*} and \begin{align*}c = 3\end{align*} into the discriminant formula \begin{align*}D = (-5)^2 - 4(1)(3) = 13\end{align*}.

There are two real solutions because \begin{align*}D > 0\end{align*}.

b) Substitute \begin{align*}a = 4, b = -4\end{align*} and \begin{align*}c = 1\end{align*} into the discriminant formula \begin{align*}D = (-4)^2 - 4(4)(1) = 0\end{align*}.

There is one real solution because \begin{align*}D = 0\end{align*}.

c) Rewrite the equation in standard form \begin{align*}-2x^2 + x - 4 = 0\end{align*}.

Substitute \begin{align*}a = -2, b = 1\end{align*} and \begin{align*}c = -4\end{align*} into the discriminant formula: \begin{align*}D= (1)^2 - 4 (-2) (-4) = -31\end{align*}.

There are no real solutions because \begin{align*}D < 0\end{align*}.

## Interpret the Discriminant of a Quadratic Equation

The sign of the discriminant tells us the nature of the solutions (or roots) of a quadratic equation. We can obtain two distinct real solutions if \begin{align*}D> 0\end{align*}, no real solutions if \begin{align*}D < 0\end{align*} or one solution (called a “double root”) if \begin{align*}D = 0\end{align*}. Recall that the number of solutions of a quadratic equation tell us how many times a parabola crosses the \begin{align*}x-\end{align*}axis.

Example 2

Determine the nature of solutions of each quadratic equation.

a) \begin{align*}4x^2 - 1 = 0\end{align*}

b) \begin{align*}10x^2 -3x = -4\end{align*}

c) \begin{align*}x^2 - 10x + 25 = 0\end{align*}

Solution

Use the value of the discriminant to determine the nature of the solutions to the quadratic equation.

a) Substitute \begin{align*}a = 4, b = 0\end{align*} and \begin{align*}c = -1\end{align*} into the discriminant formula \begin{align*}D = (0)^2 -4(4) (-1) = 16\end{align*}.

The discriminant is positive, so the equation has two distinct real solutions.

The solutions to the equation are: \begin{align*} \frac{0 \pm \sqrt{16}} {8} = \pm \frac{4} {8} = \pm \frac{1} {2}\end{align*}.

b) Rewrite the equation in standard form \begin{align*}10x^2 - 3x + 4 = 0\end{align*}.

Substitute \begin{align*}a = 10, b = -3\end{align*} and \begin{align*}c = 4\end{align*} into the discriminant formula \begin{align*}D = (-3)^2 - 4(10) (4) = -151\end{align*}.

The discriminant is negative, so the equation has two non-real solutions.

c) Substitute \begin{align*}a = 1, b = -10\end{align*} and \begin{align*}c = 25\end{align*} into the discriminant formula \begin{align*}D = (-10)^2 - 4(1) (25) = 0\end{align*}.

The discriminant is \begin{align*}0\end{align*}, so the equation has a double root.

The solution to the equation is \begin{align*} \frac{10 \pm \sqrt{0}} {2} = \frac{10} {2} = 5\end{align*}.

If the discriminant is a perfect square, then the solutions to the equation are rational numbers.

Example 3

Determine the nature of the solutions to each quadratic equation.

a) \begin{align*}2x^2 + x - 3 = 0\end{align*}

b) \begin{align*}-x^2 - 5x + 14 = 0\end{align*}

Solution

Use the discriminant to determine the nature of the solutions.

a) Substitute \begin{align*}a = 2, b = 1\end{align*} and \begin{align*}c = -3\end{align*} into the discriminant formula \begin{align*}D = (1)^2 - 4(2) (-3) = 25\end{align*}.

The discriminant is a positive perfect square so the solutions are two real rational numbers.

The solutions to the equation are \begin{align*} \frac{-1 \pm \sqrt{25}} {4} = \frac{-1 \pm 5} {4}\end{align*} so, \begin{align*}x = 1\end{align*} and \begin{align*} x = -\frac{3} {2}\end{align*}.

b) Substitute \begin{align*}a = -1, b = -5\end{align*} and \begin{align*}c = 14\end{align*} into the discriminant formula: \begin{align*}D = (-5)^2 - 4 (-1)(14) = 81\end{align*}.

The discriminant is a positive perfect square so the solutions are two real rational numbers.

The solutions to the equation are \begin{align*} \frac{5 \pm \sqrt{81}} {-2} = \frac{5 \pm 9} {-2}\end{align*} so, \begin{align*}x = -7\end{align*} and \begin{align*}x = 2\end{align*}.

If the discriminant is not a perfect square, then the solutions to the equation are irrational numbers.

Example 4

Determine the nature of the solutions to each quadratic equation.

a) \begin{align*}-3x^2 + 4x + 1 = 0\end{align*}

b) \begin{align*}5x^2 - x - 1 = 0\end{align*}

Solution

Use the discriminant to determine the nature of the solutions.

a) Substitute \begin{align*}a = -3, b = 2\end{align*} and \begin{align*}c = 1\end{align*} into the discriminant formula \begin{align*}D = (4)^2 - 4(-3)(1) = 28\end{align*}.

The discriminant is a positive perfect square, so the solutions are two real irrational numbers.

The solutions to the equation are \begin{align*} \frac{-2 \pm \sqrt{28}} {-6}\end{align*} so, \begin{align*}x \approx - 0.55\end{align*} and \begin{align*}x \approx 1.22\end{align*}.

b) Substitute \begin{align*}a = 5, b = -1\end{align*} and \begin{align*}c = -1\end{align*} into the discriminant formula \begin{align*}D = (-1)^2 - 4(5) (-1) = 21\end{align*}.

The discriminant is a positive perfect square so the solutions are two real irrational numbers.

The solutions to the equation are \begin{align*} \frac{1 \pm \sqrt{20}} {10}\end{align*} so, \begin{align*}x \approx 0.56\end{align*} and \begin{align*}x \approx -0.36\end{align*}.

## Solve Real-World Problems Using Quadratic Functions and Interpreting the Discriminant

You saw that calculating the discriminant shows what types of solutions a quadratic equation possesses. Knowing the types of solutions is very useful in applied problems. Consider the following situation.

Example 5

Marcus kicks a football in order to score a field goal. The height of the ball is given by the equation \begin{align*} y = - \frac{32} {6400} x^2 + x\end{align*} where \begin{align*}y\end{align*} is the height and \begin{align*}x\end{align*} is the horizontal distance the ball travels. We want to know if he kicked the ball hard enough to go over the goal post which is 10 feet high.

Solution

Define

Let \begin{align*}y \ =\end{align*} height of the ball in feet

\begin{align*}x \ =\end{align*} distance from the ball to the goalpost.

Translate We want to know if it is possible for the height of the ball to equal 10 feet at some real distance from the goalpost.

\begin{align*} 10 = - \frac{32} {6400} x^2 + x\end{align*}

Solve

\begin{align*}\text{Write the equation in standard form}. &&& -\frac{32} {6400} x^2 + x - 10 = 0\\ \text{Simplify}. &&& -0.005x^2 + x - 10 = 0\\ \text{Find the discriminant}. &&& D = (1)^2 - 4 (-0.005) (-10)= 0.8\end{align*}

Since the discriminant is positive, we know that it is possible for the ball to go over the goal post, if Marcus kicks it from an acceptable distance \begin{align*}x\end{align*} from the goal post. From what distance can he score a field goal? See the next example.

Example 6 (continuation)

What is the farthest distance that he can kick the ball from and still make it over the goal post?

Solution

We need to solve for the value of \begin{align*}x\end{align*} by using the quadratic formula.

\begin{align*} x = \frac{-1 \pm \sqrt{0.8}} {-0.01} \approx 10.6\ \text{or}\ 189.4\end{align*}

This means that Marcus has to be closer that 189.4 feet or further than 10.6 feet to make the goal. (Why are there two solutions to this equation? Think about the path of a ball after it is kicked).

Example 7

Emma and Bradon own a factory that produces bike helmets. Their accountant says that their profit per year is given by the function

\begin{align*}P = 0.003x^2 + 12x + 27760\end{align*}

In this equation \begin{align*}x\end{align*} is the number of helmets produced. Their goal is to make a profit of $40,000 this year. Is this possible? Solution We want to know if it is possible for the profit to equal$40,000.

\begin{align*}40000 = -0.003x^2 + 12x + 27760\end{align*}

Solve

\begin{align*}& \text{Write the equation in standard form} & & -0.003x^2 + 12x - 12240 = 0\\ & \text{Find the discriminant}. & & D = (12)^2 - 4(-0.003) (-12240) = -2.88\end{align*}

## Review Answers

1. \begin{align*}D = -24\end{align*}
2. \begin{align*}D = 57\end{align*}
3. \begin{align*}D = 0\end{align*}
4. \begin{align*}D = 1\end{align*}
5. \begin{align*}D = 384\end{align*}
6. \begin{align*}D = -95\end{align*}
7. \begin{align*}D = -15\end{align*} no real solutions
8. \begin{align*}D = 36\end{align*} two real solutions
9. \begin{align*}D = 9489\end{align*} two real solutions
10. \begin{align*}D = 0\end{align*} one real solutions
11. \begin{align*}D = -31\end{align*} no real solutions
12. \begin{align*}D = 256\end{align*} two real solutions
13. \begin{align*}D = 96\end{align*} two real irrational solutions
14. \begin{align*}D = 16\end{align*} two real rational solutions
15. \begin{align*}D = 241\end{align*} two real irrational solutions
16. \begin{align*}D = \frac{8}{3}\end{align*} two real irrational solutions
17. \begin{align*}D = 0\end{align*} one real rational solution
18. \begin{align*}D = 25\end{align*} two real rational solutions
19. no
20. yes

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