Transcript Completing the Square 1.7 What is completing the square? square?

Completing the Square 1.7
What is completing the square?
What steps do you follow to complete the
square?
Solve a quadratic equation by
finding square roots
Solve x2 – 8x + 16 = 25.
x2 – 8x + 16 = 25
Write original equation.
(x – 4)2 = 25
Write left side as a binomial
squared.
x – 4 = +5
Take square roots of each side.
x=4+5
Solve for x.
ANSWER
The solutions are 4 + 5 = 9 and 4 –5 = – 1.
Solve:
(a + b)2 =
(a - b)2 =
Make a perfect square trinomial
Find the value of c that makes x2 + 16x + c a
perfect square trinomial. Then write the
expression as the square of a binomial.
SOLUTION
STEP 1
Find half the coefficient of x. 16
2 =8
STEP 2
Square the result of Step 1. 82 = 64
STEP 3
Replace c with the result of Step 2. x2 + 16x + 64
Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2
Find the value of c that makes the expression
a perfect square trinomial. Then write the
expression as the square of a binomial.
x2 + 22x + c
SOLUTION
x
11
x2
11x
x
11 11x 121
STEP 1
22
Find half the coefficient of x.
2 = 11
STEP 2
Square the result of Step 1.
112 = 121
STEP 3
Replace c with the result of Step 2. x2 + 22x + 121
The trinomial x2 + 22x + c is a perfect square
when c = 121.
Find the value of c that makes
x2 – 6x + c
a perfect square trinomial.
b
c 
2
2
6
2

   3  9
 2 
2
Write the expression as the square of a
binomial.
x  6x  9
2
Steps for Completing the Square
ax2 + bx + c
1.Make sure the coefficient of the x2 term is
one. (If it is not, divide the equation by the
coefficient.)
2.Move the constant number to the other side.
3. Divide “b” by 2
4. Square the result from #2.
5. Add this number to both sides of the
equation.
6. Factor and solve for x.
Solve ax2 + bx + c = 0 when a = 1
Solve x2 – 12x + 4 = 0 by completing the square.
Write original equation.
x2 – 12x + 4 = 0
Write left side in the form x2 + bx.
x2 – 12x = – 4
x2 – 12x + 36 = – 4 + 36
to each side
2
(x – 6) = 32
x–6=+
x=6+
x=6+4
32
Write left side as a binomial squared.
Take square roots of each side.
Solve for x.
32
2
Simplify:
32 =
16
ANSWER
The solutions are 6 + 4 2 and 6 – 4 2
2 =4 2
Solving a Quadratic equation if
the coefficient of x2 is 1
Solve by completing the square.
x2 + 10x -3 = 0
x2 + 10x + ___ = 3 +___
x2 + 10x + 25 = 3 + 25
2
2 = 28
 10 
2
(x
+
5)
   5  25
 2
x  5   28
x  5  28
x  5  2 7
Solving a Quadratic Equation if
the Coefficient of x2 is not 1
Solve by completing the square.
3x2 – 6x + 12 = 0
3x2 − 6x + ___ = −12 + ___
3/ x2− 6/ x + ___=−12/ + ___
3
3
3
x2 −2x + ___=−4 + ___
(−2/2)2=1
x2 −2x + 1=−4 + 1
(x−1)2 = −3
x 1    3 x  1   3
x  1 i 3
Write a quadratic function in
vertex form
Write y = x2 – 10x + 22 in vertex form.
Then identify the vertex.
y = x2 – 10x + 22
y + ? = (x2 –10x + ? ) + 22
y + 25 = (x2 – 10x + 25) + 22
y + 25 = (x – 5)2 + 22
y = (x – 5)2 – 3
Write original function.
Prepare to complete the square.
2
2
–10
Add
(–5)
= 25 to each side.
2 =
Write x2 – 10x + 25 as a
binomial squared.
( )
Solve for y.
ANSWER
The vertex form of the function is y = (x – 5)2 – 3.
The vertex is (5, – 3).
Writing a Quadratic Function in
Vertex Form
Write the quadratic function in vertex
form.
y = x2 -8x +11
y + ___ = (x2 −8x + ___) +11
y + 16 = (x2 −8x + 16) +11
y + 16 = (x−4)2 +11
y = (x−4)2 −5
What is the vertex?
Find the maximum value of a quadratic function
Baseball
The height y (in feet) of a baseball t
seconds after it is hit is given by this
function:
y = –16t2 + 96t + 3
Find the maximum height of the baseball.
SOLUTION
The maximum height of the baseball is the
y-coordinate of the vertex of the parabola with the given equation.
• What is completing the square?
Another method for solving a quadratic equations.
• What steps do you follow to complete the
square?
1.Make sure the coefficient of the x2 term is one. (If
it is not, divide the equation by the coefficient.)
2.Move the constant number to the other side.
3. Divide “b” by 2
4. Square the result from #2.
5. Add this number to both sides of the equation.
6. Factor and solve for x.
Assignment 1.7
p. 54, 3-7 odd,
13-17 odd, 23-27 odd,
39-45 odd, 51