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# 5.14: Using the Discriminant

Difficulty Level: At Grade Created by: CK-12
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The profit on your school fundraiser is represented by the quadratic expression 5p2+400p8000\begin{align*}-5p^2 + 400p - 8000\end{align*}, 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=b±b24ac2a\begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}. The expression under the radical, b24ac\begin{align*}b^2-4ac\end{align*}, 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 x28x20=0\begin{align*}x^2-8x-20=0\end{align*} using the Quadratic Formula. What is the value of the discriminant?

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

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

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

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

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

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

#### Example A

Determine the type of solutions 4x25x+17=0\begin{align*}4x^2-5x+17=0\end{align*} has.

Solution: Find the discriminant.

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

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=5±252728=5±2478=58±2478i\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*}

#### Example C

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

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

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

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

This quadratic has two real solutions.

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

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

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

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 3x28x+16=0\begin{align*}-3x^2-8x+16=0\end{align*} has.

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

3. Solve the equation from #1.

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

This equation has two real solutions.

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

This equation has one real solution.

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

### Vocabulary

Discriminant
The value under the radical in the Quadratic Formula, b24ac\begin{align*}b^2-4ac\end{align*}. 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. x212x+36=0\begin{align*}x^2-12x+36=0\end{align*}
2. 5x29=0\begin{align*}5x^2-9=0\end{align*}
3. 2x2+6x+15=0\begin{align*}2x^2+6x+15=0\end{align*}
4. 6x2+8x+21=0\begin{align*}-6x^2+8x+21=0\end{align*}
5. x2+15x+26=0\begin{align*}x^2+15x+26=0\end{align*}
6. 4x2+x+1=0\begin{align*}4x^2+x+1=0\end{align*}

Solve the following equations using the Quadratic Formula.

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

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

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

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