Completing The Square - Rahway Public Schools
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Transcript Completing The Square - Rahway Public Schools
Completing The Square
Section 3.3 Beginning on page 112
The Big Idea
Completing the square is a technique we can use to write a quadratic in vertex form.
Completing the square is creating a Perfect Square Trinomial.
We would want the equation in vertex form so that we canβ¦
β’ Solve the equation using square roots.
β’ Identify the vertex (possibly to identify the maximum or minimum).
The Perfect Square Trinomial
π2 ± 2ππ + π2 = (π ± π)2
16π₯ 2 + 24π₯ + 9
= (4π₯ + 3)2
9π₯ 2 β 24π₯ + 16
= (3π₯ β 4)2
81π₯ 2 β 36π₯ + 4
= (9π₯ β 2)2
36π₯ 2 + 60π₯ + 25 = (6π₯ + 5)2
Solving Quadratics With Square Roots
Example 1: Solve π₯ 2 β 16π₯ + 64 = 100
π₯β8
2
= 100
π₯ β 8 = ±10
+8
+8
π₯ = 10 + 8
π₯ = β10 + 8
π₯ = 18
π₯ = β2
This is a perfect square
trinomial so we can factor it.
Completing The Square
If there is no perfect square trinomial, we can create something like on.
To complete the square we need to add
We must also subtract
π 2
2
π 2
2
after the b term.
to keep the equation balanced.
Example 3: Solving ππ₯ 2 + ππ₯ + π = 0 when π = 1
π
2
π₯ 2 β 10π₯ + 7 = 0
2
β10
=
2
2
= β5
π₯ 2 β 10π₯ + 25 β 25 + 7 = 0
π₯ = 5 ± 18
(π₯ β 5)2 β18 = 0
5)2 =
(π₯ β
18
π₯ β 5 = 18
+5
+5
π₯ =5± 9 2
π₯ =5±3 2
2
= 25
Completing The Square
Example 4: Solving ππ₯ 2 + ππ₯ + π = 0 when π β 1
3π₯ 2 + 12π₯ + 15 = 0
3(π₯ 2 + 4π₯ + 5) = 0
2
π₯ + 4π₯ + 5 = 0
When dividing both sides with that common factor,
you end up only with the reduced quadratic.
π
2
2
4
=
2
2
= 2
2
=4
π₯ 2 + 4π₯ + 4 β 4 + 5 = 0
(π₯ +
2)2 +1
=0
2
(π₯ + 2) = β1
π₯ + 2 = β1
β2
β2
π₯ = β2 ± β1
π₯ = β2 ± π
Writing Quadratic Functions in Vertex
Form
Example 5: Write π¦ = π₯ 2 β 12π₯ + 18 in vertex form. Then identify the vertex.
π
2
2
π¦ = π₯ β 12π₯ + 18
2
β12
=
2
2
= β6
π¦ = π₯ 2 β 12π₯ + 36 β 36 + 18
π¦ = (π₯ β 6)2 β18 = 0
The vertex is at the point (6,-18)
2
= 36
Practice
Solve the equation using square roots:
1) π₯ 2 + 4π₯ + 4 = 36
2) π₯ 2 β 6π₯ + 9 = 1
3) π₯ 2 β 22π₯ + 121 = 81
Solve the equation by competing the square:
7) π₯ 2 β 4π₯ + 8 = 0
8) π₯ 2 + 8π₯ β 5 = 0
9) β3π₯ 2 β 18π₯ β 6 = 0
10) 4π₯ 2 + 32π₯ = β68
11) 6π₯ π₯ + 2 = β42
12) 2π₯ π₯ β 2 = 200
Write the quadratic function in vertex form. Then identify the vertex:
13) π¦ = π₯ 2 β 8π₯ + 18
14) π¦ = π₯ 2 + 6π₯ + 4
15) π¦ = π₯ 2 β 2π₯ β 6
Answers:
1) π₯ = 4 πππ π₯ = β8
7) π₯ = 2 ± 2π
10) π₯ = β4 ± π
13) π¦ = (π₯ β 4)2 +2; (4,2)
2) π₯ = 4 πππ π₯ = 2
8) π₯ = β4 ± 21
11) π₯ = β1 ± π 6
14) π¦ = (π₯ + 3)2 β5; (β3, β5)
3) π₯ = 20 πππ π₯ = 2
9) π₯ = β3 ± 7
12) π₯ = 1 ± 101
15) π¦ = (π₯ β 1)2 β7; (1, β7)