1. Use the Quadratic Formula and the Discriminant Lesson 1.8
Honors Algebra 2
Use the Quadratic Formula and the Discriminant
Lesson 1.8

2. Goals
Goal
Rubric
Solve Quadratic Equations using the quadratic formula.
Use the discriminant to find the number and type of solutions to a quadratic equation.
Level 1 – Know the goals.
Level 2 – Fully understand the goals.
Level 3 – Use the goals to solve simple problems.
Level 4 – Use the goals to solve more advanced problems.
Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary
Quadratic Formula
Discriminant

4. Quadratic Formula
You have learned several methods for solving quadratic equations: completing the square, factoring, and using square roots. Another method is to use the Quadratic Formula, which allows you to solve a quadratic equation in standard form.
By completing the square on the standard form of a quadratic equation, you can derive the Quadratic Formula.

5. Deriving the Quadratic Formula

6. Axis of Symmetry
The symmetry of a quadratic function is evident in the last step,
. These two zeros are the same distance, , away from the axis of symmetry,
,with one zero on either side of the vertex.

7. Quadratic Formula
You can use the Quadratic Formula to solve any quadratic equation that is written in standard form.

8. Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.

9. Procedure: Solving a Quadratic Equation Using the Quadratic Formula
Step 1: Write the equation in standard form ax2 + bx + c = 0 and identify the values of a, b, and c.
Step 2: Substitute the values of a, b, and c into the quadratic formula.
Step 3: Simplify the expression found in Step 2.
Step 4: Check. Verify your solutions.

10. Solve x2 + 3x = 2. x2 + 3x = 2 x2 + 3x – 2 = 0 x = – b + b2 – 4ac 2a
EXAMPLE 1
Solve an equation with two real solutions
Solve x2 + 3x = 2.
x2 + 3x = 2
Write original equation.
x2 + 3x – 2 = 0
Write in standard form.
x =
– b + b2 – 4ac 2a
Quadratic formula
x =
– – 4(1)(–2)
2(1)
a = 1, b = 3, c = –2
x =
– 3 + 17 2
Simplify.
The solutions are
x =
– 3 + 17 2
0.56 and
– 3 –
– 3.56.
ANSWER

11. Solve 25x2 – 18x = 12x – 9. 25x2 – 18x = 12x – 9 25x2 – 30x + 9 = 0
EXAMPLE 2
Solve an equation with one real solution
Solve 25x2 – 18x = 12x – 9.
25x2 – 18x = 12x – 9
Write original equation.
25x2 – 30x + 9 = 0
Write in standard form.
x =
(–30)2– 4(25)(9)
2(25)
a = 25, b = –30, c = 9
x = 50
Simplify.
x = 3 5
Simplify. 3 5
The solution is
ANSWER .

12. The solutions are 2 + i and 2 – i.
EXAMPLE 3
Solve an equation with imaginary solutions
Solve –x2 + 4x = 5.
–x2 + 4x = 5
Write original equation.
–x2 + 4x – 5 = 0
Write in standard form.
x =
– – 4(– 1)(– 5)
2(– 1)
a = –1, b = 4, c = –5
x =
– – 4
– 2
Simplify.
– 4+ 2i
x =
– 2
Rewrite using the imaginary unit i.
x = 2 + i
Simplify.
The solutions are 2 + i and 2 – i.
ANSWER

13. x2 = 6x – 4 3 + 5 3 – 5 . 4x2 – 10x = 2x – 9 3 The solution is . 2
Your Turn:
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
x2 = 6x – 4
ANSWER
The solutions are
and
3 – 5 .
4x2 – 10x = 2x – 9
ANSWER 3 2
The solution is .

14. 7x – 5x2 – 4 = 2x + 3 5 + i 10 5 – i Your Turn:
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
7x – 5x2 – 4 = 2x + 3
ANSWER
The solutions are
and
115
5 + i 10
5 – i

15. The Discriminant
The expression under the radical sign in the Quadratic Formula (b2 – 4ac) is called the discriminant.
A quadratic equation can have two real solutions (x2 = 4), one real (x2 = 0), or no real solutions (x2 = -4).
The discriminant is the part of the Quadratic Formula that you can use to determine the number of real solutions of a quadratic equation.

16. The Discriminant
Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.
Caution!

17. a. x2 – 8x + 17 = 0 b. x2 – 8x + 16 = 0 c. x2 – 8x + 15 = 0 x =
EXAMPLE 4
Use the discriminant
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
a. x2 – 8x + 17 = 0
b. x2 – 8x + 16 = 0
c. x2 – 8x + 15 = 0
SOLUTION
Equation
Discriminant
Solution(s)
x =
– b+ b2– 4ac 2a
ax2 + bx + c = 0
b2 – 4ac
a. x2 – 8x + 17 = 0
(–8)2 – 4(1)(17) = – 4
Two imaginary: 4 + i
b. x2 – 8x + 16 = 0
(–8)2 – 4(1)(16) = 0
One real: 4
c. x2 – 8x + 15 = 0
(–8)2 – 4(1)(15) = 4
Two real: 3, 5

18. – 271; Two imaginary solutions
Your Turn:
for Example 4
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
2x2 + 4x – 4 = 0
48;
ANSWER
Two real solutions
3x2 + 12x + 12 = 0
0; One real solution
ANSWER
8x2 = 9x – 11
– 271; Two imaginary solutions
ANSWER

19. – 108; Two imaginary solutions
Your Turn:
for Example 4
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
7x2 – 2x = 5
144; Two real solutions
ANSWER
4x2 + 3x + 12 = 3 – 3x
– 108; Two imaginary solutions
ANSWER
3x – 5x2 + 1 = 6 – 7x
0; One real solution
ANSWER

20. EXAMPLE 5
Solve a vertical motion problem
Juggling
A juggler tosses a ball into the air. The ball leaves the juggler’s hand 4 feet above the ground and has an initial vertical velocity of 40 feet per second. The juggler catches the ball when it falls back to a height of 3 feet. How long is the ball in the air?
SOLUTION
Because the ball is thrown, use the model h = –16t2 + v0t + h0. To find how long the ball is in the air, solve for t when h = 3.

21. h = –16t2 + v0t + h0 3 = –16t2 + 40t + 4 0 = –16t2 + 40t + 1
EXAMPLE 5
Solve a vertical motion problem
h = –16t2 + v0t + h0
Write height model.
3 = –16t2 + 40t + 4
Substitute 3 for h, 40 for v0, and 4 for h0.
0 = –16t2 + 40t + 1
Write in standard form.
t =
– – 4(– 16)(1)
2(– 16)
Quadratic formula
t = –
– 32
Simplify.
t – or t
Use a calculator.

22. EXAMPLE 5
Solve a vertical motion problem
ANSWER
Reject the solution – because the ball’s time in the air cannot be negative. So, the ball is in the air for about 2.5 seconds.

23. The ball is in the air for about 3.1 seconds.
Your Turn:
for Example 5
WHAT IF? In Example 5, suppose the ball leaves the juggler’s hand with an initial vertical velocity of 50 feet per second. How long is the ball in the air?
The ball is in the air for about 3.1 seconds.
ANSWER