Pre-Calculus 11 Updated January 2020 Adrian Herlaar, School District 61 www.mrherlaar.weebly.com Section 4.3 – Solving Quadratics by Factoring and the Square Root Method This Booklet Belongs to: Block: Definition of a Quadratic Equation An equation that can be written in the form: + + = Where , , are Real Numbers with ≠0 The solutions of the quadratic equation () = 2 + + = 0 are called zeros, roots, or solutions The points where the graph crosses the − are real solutions, because () is The − are called the real roots of the quadratic function A quadratic function can cross the − either , , Real Zeros of a Quadratic Function If () is a quadratic function and is a real zero of (), then the following statements are equivalent 1. = ℎ () 2. = ℎ () 3. = ℎ () = 0 4. ( − ) ℎ () 5. (, 0) − ℎ ℎ () Example 1: () = 2 − 2 + 3 () = 2 − 2 + 1 () = 2 − 2 − 3 No zeros (No real roots) One zero (Double root/two roots equal) Two zeros (Two unequal real roots)
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Pre-Calculus 11 Updated January 2020
Adrian Herlaar, School District 61 www.mrherlaar.weebly.com
Section 4.3 – Solving Quadratics by Factoring and the Square Root Method
This Booklet Belongs to: Block:
Definition of a Quadratic Equation An equation that can be written in the form:
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎 Where 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 are Real Numbers with 𝑎 ≠ 0
The solutions of the quadratic equation 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 are called zeros, roots, or solutions
The points where the graph crosses the 𝒙 − 𝒂𝒙𝒊𝒔 are real solutions, because 𝒇(𝒙) is 𝟎
The 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔 are called the real roots of the quadratic function
A quadratic function can cross the 𝑥 − 𝑎𝑥𝑖𝑠 either 𝟎, 𝟏, 𝒐𝒓 𝟐 𝒕𝒊𝒎𝒆𝒔
Real Zeros of a Quadratic Function If 𝑓(𝑥) is a quadratic function and 𝑐 is a real zero of 𝑓(𝑥), then the following statements are equivalent
Adrian Herlaar, School District 61 www.mrherlaar.weebly.com
Example: Solve the equation 𝑥(3𝑥 + 1) = 2
Solution:
Example: 𝑥
𝑥−5−
3
𝑥+1=
30
𝑥2−4𝑥−5
Solution:
𝑥(3𝑥 + 1) = 2 → 3𝑥2 + 𝑥 = 2
3𝑥2 + 𝑥 − 2 = 0 → 𝑥2 + 𝑥 − 6 = 0 → (𝑥 + 3)(𝑥 − 2)
(𝑥 +3
3) (𝑥 −
2
3) = 0 → 𝑥 = −
3
3= −1 𝑜𝑟 𝑥 =
2
3
𝑥
𝑥 − 5−
3
𝑥 + 1=
30
𝑥2 − 4𝑥 − 5
𝑥
𝑥 − 5−
3
𝑥 + 1=
30
(𝑥 − 5)(𝑥 + 1)
(𝑥 − 5)(𝑥 + 1) [𝑥
𝑥 − 5−
3
𝑥 + 1=
30
(𝑥 − 5)(𝑥 + 1)]
(𝑥 + 1)𝑥 − 3(𝑥 − 5) = 30
𝑥2 + 𝑥 − 3𝑥 + 15 − 30 = 0
𝑥2 − 2𝑥 − 15 = 0 → (𝑥 − 5)(𝑥 + 3)
𝒙 = 𝟓 𝒐𝒓 𝒙 = −𝟑
When finished inset your answers into the original equation to make sure they are valid
The denominator cannot equal zero when you plug it in
So it this case reject 𝒙 = 𝟓
Factor the denominators
Identify LCM
In this case: (𝑥 − 5)(𝑥 + 1)
Multiply each term by the
LCM
Reduce
Simplify
Factor
Solve
Pre-Calculus 11 Updated January 2020
Adrian Herlaar, School District 61 www.mrherlaar.weebly.com
Solving Quadratics Using the Square Root Method
The factor method is definitely the most efficient method of solving quadratics
But not all quadratics can be factored easily of at all
What we get are two methods depending on the situation
The Square Root Method
The Quadratic Equation
The square root method in mainly used when 𝑏 = 0 in the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
Isolate the 𝑥2 on the left side and square root the other side
Example:
𝑥2 − 16 = 0 𝑥2 − 16 = 0
(𝑥 + 4)(𝑥 − 4) or 𝑥2 = 16
𝑥 − 4 = 0 𝑜𝑟 𝑥 + 4 = 0 √𝑥2 = ±√16
𝑥 = 4 𝑜𝑟 𝑥 = −4 𝑥 = ±4
The procedure on the right is the SQUARE ROOT METHOD
Solving a Quadratic Equation of the form 𝒂𝒙𝟐 + 𝒄 = 𝟎 Step 1: Isolate the 𝒙𝟐 on the left side of the equation and the constant on the right Step 2: Take the square root of both sides, the square root of the constant has to be ± Step 3: Simplify if possible Step 4: Check the solution in the original equation
The Square Root Property is defined as follows:
The Square Root Property The equation 𝑥2 = 𝑛 has exactly 2 real solutions