Solving Quadratic Equations

Solving Quadratic Equations Notes

Today's Goal

• ^{Be able to solve for "x" in a quadratic equation by factoring.}

• ^{Understand that when we are solving quadratic equations, we are finding x-values for when we know "y".}

• ^{Understand why x-intercepts are called solutions or zeros.}

• Factoriza la ecuación cuadrática.

• Toma un binomio o factor a la vez y resuelve esa "x".

• Tendrás dos, una o ninguna solución.

• Factor the quadratic equation.

• Take one binomial or factor at a time and solve for that "x".

• You will have two, one, or no solutions.

Steps to Solve:

Example: x² + 5x + 4 = 0

1. Factor the equation (leave =0 alone):

   x² + 5x + 4 -> (x + __) (x + __)

   *Two numbers that multiply to make 4 but add to make 5.

Multiple Select

What two numbers multiply to make 4 but add to make 5? (Choose 2)

• 1

• 2

• 4

• 0

Example: x² + 5x + 4 = 0

1. Factor the equation (leave =0 alone):

   x² + 5x + 4 -> (x + 1) (x + 4)

2. Take each binomial or factor and solve.

   x + 1 = 0 x = ?

   x + 4 = 0 x = ?

Multiple Choice

Solve for x:

x + 1 = 0

• 1

• 0

• 2

• -1

Example: x² + 5x + 4 = 0

1. Factor the equation (leave =0 alone):

   x² + 5x + 4 -> (x + 1) (x + 4)

2. Take each binomial or factor and solve.

   x + 1 = 0 x = -1

   x + 4 = 0

Multiple Choice

Solve:

x + 4 = 0

• 4

• -4

• 0

• 1

Example: x² + 5x + 4 = 0

1. Factor the equation (leave =0 alone):

   x² + 5x + 4 -> (x + 1) (x + 4)

2. Take each binomial or factor and solve.

   x + 1 = 0 x = -1

   x + 4 = 0 x = -4

Example: x² + 5x + 4 = 0

x² + 5x + 4 -> (x + 1) (x + 4)

x + 1 = 0 x = -1

x + 4 = 0 x = -4

Answer: x = -1 and -4

This means, when y is 0 (on the x-axis), our parabola will cross at -1 and -4.

Example:
x² + 7x + 12 = 0

Multiple Choice

Factor x² + 7x + 12

• (x + 3)(x + 4)

• (x + 1)(x + 12)

• (x + 2)(x + 6)

• (x + 1)(x + 7)

Example:
x² + 7x + 12 = 0 ->
(x + 3)(x + 4) = 0

Multiple Choice

Solve:

x + 3 = 0

• x = 3

• x = -3

• x = 0

• x = 1/3

Multiple Choice

Solve:

x + 4 = 0

• x = -4

• x = 4

• x = 1/2

• x = 0

Example:
x² + 7x + 12 = 2 ->
(x + 3)(x + 4) = 0
x = -3, -4

Multiple Choice

Step 1: Factor

x² - 9x + 14 = 0

• (x - 1)(x - 14) = 0

• (x + 1)(x + 14) = 0

• (x + 2)(x + 7) = 0

• (x - 2)(x - 7) = 0

Multiple Select

Step 2 & 3: Separate and Solve (Pick 2)

x² - 9x + 14 = 1

(x - 2)(x - 7) = 1
x - 2 = 1 ; x - 7 = 1

• x = 2

• x = 7

• x = 3

• x = 8

• x = -9

Multiple Choice

Factor:

x² + 7x + 10

• (x + 10)(x + 1)

• (x + 2)(x + 5)

• (x + 10)(x + 0)

• (x + 7)(x + 3)

Multiple Choice

x² +7x + 10 = 0

(x + 2)(x + 5) = 0
Find the solution(s) to x

• x = -2, -5

• x = 7, 10

• x = 3, 0

• x = -3, 0

What happens if x = 0?

If x = 0, it counts as a solution!

Number of Solutions

• 2 Solutions: We are able to factor and have 2 numbers that are different.

  • Example: x² + 3x + 2 = (x + 1)(x + 2)

• 1 Solution: We are able to factor, but they are exactly the same.

  • Example: x² + 10x + 25 = (x + 5)(x + 5) or (x + 5)²

• No solutions: The equation is not factorable/No x-intercepts

  • Example: x² + 2x + 3 = Not factorable/No x-intercepts