Warm-up 1. Solve the following quadratic equation by Completing the Square: 2x2 - 20x + 16 = 0 2. Convert the following quadratic equation to vertex.

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Warm-up 1. Solve the following quadratic equation by Completing the Square: x x + 15 = 0 2. Convert the following quadratic equation to vertex format.
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Presentation transcript:

Warm-up 1. Solve the following quadratic equation by Completing the Square: 2x2 - 20x + 16 = 0 2. Convert the following quadratic equation to vertex format y = 3x2 – 12x + 23

Section 4-8 The Quadratic Formula Chapter 4 Section 4-8 The Quadratic Formula

Objectives I can use the Quadratic Formula to solve for the solutions of any quadratic equation

Quadratic Review Quadratic Equation in standard format: y = ax2 + bx + c Solutions (roots) are where the graph crosses or touches the x-axis. Solutions can be real or imaginary

Types of Solutions 2 Real Solutions 1 Real Solution 2 Imaginary Solutions

Key Concept for this Section What happens when you square any number like below: x2 = ? It is always POSITIVE!! This is always the biggest mistake in this section

Key Concept #2 What happens when you subtract a negative number like below: 3 - -4 = ? It becomes ADDITION!! This is 2nd biggest error on this unit!

The Quadratic Formula The solutions of any quadratic equation in the format ax2 + bx + c = 0, where a  0, are given by the following formula: x = The quadratic equation must be set equal to ZERO before using this formula!!

Example 1 x2 – 7x = 18 First set equation equal to 0 x2 – 7x – 18 = 0 Now identify a, b, and c a = 1, b = -7, c = -18 Now plug into formula x = = x = = x = = 18/2 0r -4/2 x = 9 or -2

Your Turn Solve the following in your notes: -3x2 + 4x – 4 = 0 Roots:

Another Attempt Solve the following in your notes: x2 + 9x – 11 = 0 Roots:

Solve an equation with two real solutions 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 = – 3 + 32 – 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

Solve an equation with imaginary solutions 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+ 42– 4(– 1)(– 5) 2(– 1) a = –1, b = 4, c = –5 x = – 4+ – 4 – 2 Simplify. – 4+ 2i x = – 2 Rewrite using the imaginary unit i. x = 2 + i Simplify. The solution is 2 + i and 2 – i. ANSWER

Homework WS 6-4 Factoring Quiz