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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 $-5p^2 + 400p - 8000$ , where p is your price point. How many real break-even points will you have?

### Guidance

From the previous concept, the Quadratic Formula is $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . The expression under the radical, $b^2-4ac$ , is called the discriminant. You can use the discriminant to determine the number and type of solutions an equation has.

#### Investigation: Solving Equations with Different Types of Solutions

1. Solve $x^2-8x-20=0$ using the Quadratic Formula. What is the value of the discriminant?

$x &=\frac{8 \pm \sqrt{{\color{red}144}}}{2}\\&=\frac{8 \pm 12}{2} \rightarrow 10, -2$

2. Solve $x^2-8x+6=0$ using the Quadratic Formula. What is the value of the discriminant?

$x &=\frac{8 \pm \sqrt{{\color{red}0}}}{2}\\&=\frac{8 \pm 0}{2} \rightarrow 4$

3. Solve $x^2-8x+20=0$ using the Quadratic Formula. What is the value of the discriminant?

$x &=\frac{8 \pm \sqrt{{\color{red}-16}}}{2}\\&=\frac{8 \pm 4i}{2} \rightarrow 4 \pm 2i$

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

From this investigation, we can conclude:

• If $b^2-4ac > 0$ , then the equation has two real solutions.
• If $b^2-4ac = 0$ , then the equation has one real solution; a double root.
• If $b^2-4ac < 0$ , then the equation has two imaginary solutions.

#### Example A

Determine the type of solutions $4x^2-5x+17=0$ has.

Solution: Find the discriminant.

$b^2-4ac &=(-5)^2-4(4)(17)\\&=25-272$

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.

#### Example B

Solve the equation from Example A to prove that it does have two imaginary solutions.

$x = \frac{5 \pm \sqrt{25-272}}{8}=\frac{5 \pm \sqrt{-247}}{8}= \frac{5}{8} \pm \frac{\sqrt{247}}{8}i$

#### Example C

Find the value of the determinant and state how many solutions the quadratic has.

$3x^2-5x-12=0$

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

$\sqrt{(-5)^2 - 4(3)(-12)}=\sqrt{25+144}=\sqrt{169}=13$

This quadratic has two real solutions.

Intro Problem Revisit Set the expression $-5p^2 + 400p - 8000$ equal to zero and then find the discriminant.

$-5p^2 + 400p - 8000 = 0$

$b^2-4ac &=(400)^2-4(-5)(-8000)\\&=160000-160000 = 0$

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

### Guided Practice

1. Use the discriminant to determine the type of solutions $-3x^2-8x+16=0$ has.

2. Use the discriminant to determine the type of solutions $25x^2-80x+64=0$ has.

3. Solve the equation from #1.

1. $b^2-4ac &=(-8)^2-4(-3)(16)\\&=64+192\\&=256$

This equation has two real solutions.

2. $b^2-4ac &=(-80)^2-4(25)(64)\\&=6400-6400\\&=0$

This equation has one real solution.

3. $x = \frac{8 \pm \sqrt{256}}{-6}=\frac{8 \pm 16}{-6}=-4, \frac{4}{3}$

### Vocabulary

Discriminant
The value under the radical in the Quadratic Formula, $b^2-4ac$ . The discriminant tells us number and type of solution(s) a quadratic equation has.

### Practice

Determine the number and type of solutions each equation has.

1. $x^2-12x+36=0$
2. $5x^2-9=0$
3. $2x^2+6x+15=0$
4. $-6x^2+8x+21=0$
5. $x^2+15x+26=0$
6. $4x^2+x+1=0$

Solve the following equations using the Quadratic Formula.

1. $x^2-17x-60=0$
2. $6x^2-20=0$
3. $2x^2+5x+11=0$

Challenge Determine the values for $c$ that make the equation have a) two real solutions, b) one real solution, and c) two imaginary solutions.

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