First Quarter - Chapter 2 - Quadratic Equation

Quadratic Equation

1 – Definition of Quadratic Equation2 – Solving

Quadratic Equations by the Square Root Property3

– Solving Quadratic Equations by Completing the Square

4 – Solving Quadratic Equations by the Quadratic Formula

5 – Graphing Quadratic Equations in Two Variables6 –

Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions

*Sections

*Quadratic Equations*An example of a Quadratic Equation:

*The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x2). *It is also called an "Equation of Degree 2" (because of the "2" on the x)

The letters a, b and c are coefficients (you know those values).

They can have any value, except that a can't be 0. The letter "x" is the variable or unknown

(you don't know it yet)

The Standard Form of a Quadratic Equation looks like this:

Solving Quadratic Equations by the Square Root Property

*Square Root PropertyWe previously have used factoring to solve quadratic equations.This chapter will introduce additional methods for solving quadratic equations.Square Root Property

If b is a real number and a2 = b, then

ba

Solve x2 = 49

2xSolve (y – 3)2 = 4

Solve 2x2 = 4x2 = 2

749 x

y = 3 2 y = 1 or 5

243 y

Example

Solve x2 + 4 = 0 x2 = 4

There is no real solution because the square root of 4 is not a real number.

Example

Solve (x + 2)2 = 25

x = 2 ± 5 x = 2 + 5 or x = 2 – 5x = 3 or x = 7

5252 x

Example

Solve (3x – 17)2 = 28

72173 x

37217

x

7228 3x – 17 =

Example

Solving Quadratic Equations by Completing the Square

In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left.Also, the constant on the left is the square of the constant on the right.So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).

Completing the Square

What constant term should be added to the following expressions to create a perfect square trinomial?x2 – 10x

add 52 = 25x2 + 16x

add 82 = 64x2 – 7xadd 4

4927 2

Example

We now look at a method for solving quadratics that involves a technique called completing the square.It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.

Example

Solving a Quadratic Equation by Completing a Square

1) If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient.

2) Isolate all variable terms on one side of the equation.

3) Complete the square (half the coefficient of the x term squared, added to both sides of the equation).

4) Factor the resulting trinomial.5) Use the square root property.

Solve by completing the square. y2 + 6y = 8y2 + 6y + 9 = 8 + 9 (y + 3)2 = 1

y = 3 ± 1 y = 4 or 2

y + 3 = ± = ± 1

1

Solving EquationsExampl

e

Solve by completing the square. y2 + y – 7 = 0 y2 + y = 7

y2 + y + ¼ = 7 + ¼

229

429

21

y

2291

229

21

y

(y + ½)2 = 429

Example

Solve by completing the square. 2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½

251

451

27

x2

517251

27

x

x2 + 7x + = ½ + = 4

494

49451

(x + )2 = 451

27

Example

Solving Quadratic Equations by the Quadratic Formula

*The Quadratic Formula

Another technique for solving quadratic equations is to use the quadratic formula.

The formula is derived from completing the square of a general quadratic equation.

A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.

aacbbx

242

Solve 11n2 – 9n = 1 by the quadratic formula.11n2 – 9n – 1 = 0, soa = 11, b = -9, c = -1

)11(2

)1)(11(4)9(9 2

n

2244819

221259

22559

Example

)1(2

)20)(1(4)8(8 2

x

280648

21448

2

128 20 4 or , 10 or 22 2

x2 + 8x – 20 = 0 (multiply both sides by 8)a = 1, b = 8, c = 20

81

25Solve x2 + x – = 0 by the

quadratic formula.

Example

Solve x(x + 6) = 30 by the quadratic formula.x2 + 6x + 30 = 0

a = 1, b = 6, c = 30

)1(2

)30)(1(4)6(6 2

x

2120366

2846

So there is no real solution.

Example

The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant.The discriminant will take on a value that is positive, 0, or negative.The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

*The Discriminant

Use the discriminant to determine the number and type of solutions for the following equation.

5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5

b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224

There are no real solutions.

Example

3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16

b2 – 4ac b2 – 4ac b2 – 4ac

(–2)2 – 4(3)(2) 112 – 4(2)(12) 82 – 4(1)(16)

4 – 24 121 – 96 64 – 64

–20 25 0

b2 – 4ac is negative.There are no real

solutions

b2 – 4ac is positive.There are two real

solutions

b2 – 4ac is zero.There is one real

solution

Example 2: Using the Discriminant

Find the number of solutions of each equation using the discriminant.

A. B. C.

Check It Out! Example 2Find the number of solutions of each equation using the discdriminant.

2x2 – 2x + 3 = 0 x2 + 4x + 4 = 0 x2 – 9x + 4 = 0a = 2, b = –2, c = 3 a = 1, b = 4, c = 4 a = 1, b = –9 , c = 4

b2 – 4ac b2 – 4ac b2 – 4ac

(–2)2 – 4(2)(3) 42 – 4(1)(4) (–9)2 – 4(1)(4)

4 – 24 16 – 16 81 – 16

–20 0 65

b2 – 4ac is negative.There are no real

solutions

b2 – 4ac is positive.There are two real

solutions

b2 – 4ac is zero.There is one real

solution

a. b. c.

The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t2 + vt + c, where c is the initial height of the object above the ground. The ringer on a carnival strength test is 2 feet off the ground and is shot upward with an initial velocity of 30 feet per second. Will it reach a height of 20 feet? Use the discriminant to explain your answer.

Application

Continuedh = –16t2 + vt + c

20 = –16t2 + 30t + 2

0 = –16t2 + 30t + (–18)

b2 – 4ac

302 – 4(–16)(–18) = –252

Substitute 20 for h, 30 for v, and 2 for c.

Subtract 20 from both sides.

Evaluate the discriminant.

Substitute –16 for a, 30 for b, and –18 for c.

The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

Check It Out! Example 2 What if…? Suppose the weight is shot straight up with an initial velocity of 20 feet per second from 1 foot above the ground. Will it ring the bell? Use the discriminant to explain your answer.

h = –16t2 + vt + c

20 = –16t2 + 20t + 1

0 = –16t2 + 20t + (–19)

b2 – 4ac

202 – 4(–16)(–19) = –816

Substitute 20 for h, 20 for v, and 1 for c.

Subtract 20 from both sides.

Evaluate the discriminant.

Substitute –16 for a, 20 for b, and –19 for c.

The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.