Essential Question: What are some things the discriminate is used for?

Basic strategy◦ Add or subtract the same quantity from both

sides of the equation◦ Multiply or divide both sides of the equation by

the same nonzero quantity. Definition of a Quadratic Equation

◦ A quadratic, or second degree, equation is one that can be written in the form: ax2 + bx + c = 0

◦ For real constants a, b, and c, with a ≠ 0

Techniques used to solve quadratic equations◦ There are four techniques used to algebraically

find exact solutions of quadratic equations.◦ Techniques that can be used to solve some

quadratic equations:1. Factoring2. Taking the square root of both sides of an equation

◦ Techniques that can be used to solve all quadratic equations3. Completing the square4. Using the quadratic formula

Factoring1. Rearrange the terms so that everything equals 02. Find two numbers that multiply together to get a•c &

add together to get b3. Use those numbers to split b (the term in the middle)4. Take out the greatest common factor in each group5. Group the outside terms together6. Set each part equal to 0 and solve

Example 1: Solve 3x2 – x = 10 by factoring 1) 3x2 – x – 10 = 0 [ax2 + bx + c = 0]2) Numbers that multiply to get -30, add to get -1?

-6 and 53) (3x2 – 6x) + (5x – 10) = 04) 3x(x – 2) + 5(x – 2) = 05) (3x + 5)(x – 2) = 06) x = -5/3 or x = 2

Taking the square root of both sides◦ Only works when x2 = k

(a positive constant - no “bx” term)◦ For a real number k:

Number of Solutions

Solutions

k < 0 0

k = 0 1 0

k > 0 2 and k k

Taking the square root of both sides◦ Example 2: Solve 3x2 = 91.Get squared term by itself

x2 = 3

2.Take the square root of both sides x = 3

Taking the square root of both sides◦ Example 3: Solve 2(x + 4)2 = 6

1.Get squared term by itself (x + 4)2 = 3

2.Take the square root of both sides x + 4 =

3.Get x by itself x = -4

3

3

Assignment◦ Page 95, 1-23 (odds)

Don’t expect to get credit without showing work

Essential Question: What are some things the discriminate is used for?

Note: Completing the square is really only useful for determining the quadratic equation. We’ll do that after this short demonstration…

Completing the square1. Write the equation in the form x2 + bx = c

2. Add to both sides so that the left side is a

perfect square and the right side is a constant3. Take the square root of both sides4. Simplify

2

2

b

Completing the square◦ Example 4: Solve 2x2 – 6x + 1 = 0 by completing the

square1.Write the equation in the form x2 + bx = c

2x2 – 6x = -1 x2 – 3x = -½

2.Add to both sides so that the left side is a perfect

square and the right side is a constant

2

2

b

2

2

9 1 93

4 2 4

3 7

2 4

x x

x

Completing the square (continued)

1. Take the square root of both sides

2. Simplify 3 7

2 4x

3 7

2 4x

23 7

2 4x

The Quadratic Formula

◦ The solutions of the quadratic equation

ax2 + bx + c = 0 are

◦ Get an equation to equal 0, then simply substitute in the formula

2 4

2

b b acx

a

Solve x2 + 3 = -8x by using the quadratic formula◦ Get equation to equal 0

x2 + 8x + 3 = 0◦ Plug into the quadratic formula

a = 1, b = 8, c = 3

22 8 8 4 1 34

2 2 1

8 64 12 8 52

2 2

8 4 13 8 2 134 13

2 2

b b ac

a

The Discriminant◦ The portion of the quadratic formula that exists

underneath the square root (b2 – 4ac) is called the discriminant. It can be used to determine the number of real solutions of a quadratic equation.

Discriminant Value Number of Real Solutions ofax2 + bx + c = 0

b2 – 4ac < 0 0 real solutions

b2 – 4ac = 0 1 distinct real solution

b2 – 4ac > 0 2 distinct real solutions

Using the discriminant◦ Find the number of real solutions to 2x2 = -x – 31.Write the equation in general form

(make one side = 0) 2x2 + x + 3 = 0

2.Plug into the discriminant and simplify a = 2, b = 1, c = 3 b2 – 4ac (1)2 – 4(2)(3) = 1 – 24 = -23

◦ Because -23 < 0, the equation has no real solutions

◦ We’ll confirm using the calculator

Polynomial Equations◦ A polynomial equation of degree n is an equation that

can be written in the form: anxn + an-1xn-1 + … + + a1x + a0 = 0

◦ Example: 4x6 – 3x5 + x4 + 7x3 – 8x2 + 4x + 9 = 0 is a polynomial equation of degree 6.

◦ Example 2: 4x3 – 3x2 + 4x - 5 = 0 is a polynomial expression of degree 3.

◦ Polynomials have the following traits No variables in denominators (integers only) No variables under radical signs

◦ Polynomials of degree 3 and above are best solved graphically. However, some equations are quadratic in form and can be solved algebraically.

Polynomial Equations in Quadratic Form◦ Solve 4x4 – 13x2 + 3 = 0◦ To solve, we substitute u for x2

4u2 – 13u + 3 = 0◦ Then solve the quadratic equation

(4u – 1)(u – 3) = 0 u = ¼ or u = 3

◦ Because u = x2:

2 21 or 3

4

1 1 or 3

4 2

x x

x x

Assignment◦ Page 95-96, 25-53 (odds)

For the problems that direct you to solve by completing the square, use the quadratic formula instead.

Hints for 47-53 47, 49 & 51 can be factored (though you don’t have to

solve by factoring) Don’t expect to get credit without showing work