ax2 +bx+c=0 x2 − 8x + 15 = 0 (x − 5)( x − 3) = 0 x − 5 = 0 x = 5 x − 3

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4.4 NON LINEAR EQUATIONS p73 textbook
Quadratic Equation
2
- the general form of the quadractic equation is written as ax +bx +c = 0
- a polynomial equation of second degree
- has at most two solutions or roots
Eg)
where a ! 0
factor
x 2 − 8x + 15 = 0
(x − 5 )(x − 3 ) = 0
The equation requires that the contents of at least one of the brackets will equal 0
- what values of x will satisfy these conditions
x−5=0
x=5
x−3=0
x=3
- 5 and 3 are the roots of this quadratic equation and this can be interpreted
as the x axis intercepts
plot of y = x 2 − 8 x + 15
20
15
10
5
0
2
4
x
6
8
10
-5
- note that the quadratic crosses the x axis at x = 3 and x = 5
- note that the y axis intercept is 15, when x = 0, y = 15
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Eg)
solve or find the roots of
x 2 + 5x = 0
x(x + 5 ) = 0
x=0
x+5=0
Eg)
x = −5
find the roots or solve
7x 2 − 3 = 0
7x 2 = 3
x2 = 3
7
x=! 3
7
Eg)
solve
5x 2 = 15x
5x 2 − 15x = 0
5x(x − 3 ) = 0
x=0
x−3=0
x=3
Eg)
solve
16x 2 − 1 = 0
= (4x − 1 )(4x + 1 )
x = !1
4
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Eg)
Solve skip this one
x+ x−4 = 4
x−4 = 4−x
x − 4 = (4 − x) 2 = 16 − 8x + x 2
x 2 − 9x + 20 = 0
(x − 5)(x − 4) = 0
x = 5, 4
-however x = 4 is an extraneous root
- factoring up to this point involved integer or whole number factors
Eg)
factor
x 2 + 4x − 2
= (x )(x )?????
by inspection
−b ! b 2 − 4ac
x=
2a
- apply the quadratic formula
x 2 + 4x − 2 = 0
a = 1 b = 4 c = −2
20
15
−4 ! 4 2 − 4 % 1 % (−2 )
2%1
−4 ! 24
=
2
= −4 ! 4.899
2
= −4.450, +0.449
10
x=
5
-5
-4
-3
x
-2
-1
0
1
2
-5
x 2 + 4x − 2
= (x + 4.450 )(x − 0.449 ) = 0
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3
Eg)
solve
− x 2 − 3x + 2 = 0
a = −1 b = −3 c = 2
x=
3 ! 3 2 − 4 % (−1 ) % 2
2 % (−1 )
4
2
-5
-4
-3
x
-2
-1
3! 9+8
−2
3 − 17 3 + 17
=
,
−2
−2
= −3.562, +0.562
2
3
-2
=
Eg)
1
0
-4
-6
-8
-10
solve
x 2 − 4x + 5 = 0
a = 1 b = −4 c = 5
x=
4±
( −4 )
2
− 4 × 1× 5
2 ×1
4 ± −4
imaginary solutlion
2
4 ± −4 4 ± 4 × −1 4 ± 4 −1 4 ± 2i
=
=
=
=
= 2±i
2
2
2
2
=
- since the square root of a negative number is not real, there are no real solutions for x
10
8
6
4
2
-2
0
2
4
x
6
8
-2
- no real solutions means that there are no crossings of the x axis
Homework textbook pg 78 1, 3, 5, 7, 9
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The Discriminant can be used to test for real solutions
D = b 2 − 4ac
15
- if D < 0 then there are no real solutions
- if D = 0, then there is one real solution
- if D > 0 then there are two real solutions
Eg)
10
5
solve
-5
x 2 + 3x − 2 = 0
-4
-3
x
-2
0
-1
1
2
3
-5
a = 1 b = 3 c = −2
−3 ! 3 2 − 4 % 1 % (−2 )
x=
2%1
−3 ! 17
=
2
= −4 ! 4.123
2
= −8.123, +0.123
Eg)
solve
3x 2 + 5x − 2 = 0
a = 3 b = 5 c = −2
20
x=
−5 ± 52 − 4 × 3 × ( −2 )
2×3
−5 ± 49 −5 ± 7
1
=
=
= −2,
6
6
3
15
10
5
-4
-3
-2
-1
0
-5
Demonstrate the use of the Sharp EL546W calculator for solving quadractic equations
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1
x
2
Eg)
solve
3x 2 + 5x + 10 = 0
50
a = 3 b = 5 c = 10
40
−5 ! 5 2 − 4 % 3 % 10
x=
2%3
−5 ! −95
=
6
30
20
10
-4
-2
0
2
4
x
- no real roots
Eg)
solve
x2 − 2 = 0
10
8
a = 1 b = 0 c = −2
6
4
0 ! 0 2 − 4 % 1 % (−2 )
2%1
! 8
=
2
= !1.414
2
x=
Eg)
-3
-2
-1
0
1
2
x
3
-2
solve
2x 2 = 0
30
a=2 b=0 c=0
20
02 ! 0 − 4 % 2 % 0
2%2
= 0 =0
4
10
x=
-4
-2
0
2
-10
Homework text p 78 11, 13, 15, 17, 19
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x
4
Sketching the Quadratic
- the second degree polynomial produces a parabola
- a sketch can be made by determining the following
y orientation
y Y-intercept
y X-intercept
y vertex
y
Eg)
25
sketch
20
y = 2x − 5x − 6
2
15
10
y = ax 2 + bx + c
5
0
-2
a = 2, b = −5, c = −6
0
2
-5
Graph paper printing software can be downloaded from
http://perso.easynet.fr/~philimar/graphpapeng.htm
y = ax 2 + bx + c
a > 0 Opens up
a < 0 Opens down
Curve orientation
( 0, c )
Y-axis intercept
X-axis intercepts
Solve for the roots
x=
vertex
−b! b 2 −4ac
2a
x = −b
2a
- since a > 0, then the curve opens up
- y-axis intercept when x = 0
y = 2(0) 2 − 5 % 0 − 6 = −6 (0, −6)
- x-axis intercepts
2
5 ! (−5 ) − 4 % 2 % (−6)
5 ! 73
x=
=
2%2
4
x = −0.886, 3.386
- vertex
−(−5) 5
x = −b =
= = 1.25
2a 2 % 2 4
y = 2(1.25 ) 2 − 5(1.25 ) − 6 = −9.125
- by symmetry at y = -6, x = 2 X 1.25 = 2.5
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(1.25, −9.125)
( 2.5, -6 )
11/10/2007
4
x
Eg)
sketch
y = −1x 2 + 8x − 16
y = ax 2 + bx + c
a = −1, b = 8, c = −16
x
-2.5
0
2.5
5
7.5
10
0
-12.5
- since a < 0, then the curve opens down
-25
-37.5
- y-axis intercept when x = 0
y = −1(0) 2 + 8 % 0 − 16 = −16 (0, −16)
-50
y
- x-axis intercepts
−8 ! 8 2 − 4 % (−1) % (−16)
−8 ! 0
x=
=
−2
2 % (−1)
x=4
- there is only one root - meaning that the curve is tangent to the x-axis at x = 4
- vertex
x = −b = −8 = 4
2a 2 % (−1)
y = −1(4 ) 2 + 8(4 ) − 16 = 0
By symmetry x = 2 X 4 = 8
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(4, 0)
( 8, -16 )
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Eg)
sketch
y = 3x 2 − 5x + 4 = 0
y
a = 3, b = −5, c = 4
5
3.75
2.5
- since a > 0, then the curve opens up
1.25
0
-0.5
0
0.5
1
1.5
2
2.5
x
- y-axis intercept when x = 0
y = 3(0) 2 − 5 % 0 + 4 = 4
(0, 4)
- x-axis intercepts
5 ! 52 − 4 % 3 % 4
5 ! −23
=
2%3
6
x=4
x=
- no real roots, the curve does not intersect the x axis
- vertex
x = −b = 5 = 5 = 0.833
2a 2 % 3 6
2
5
− 5 5 + 4 = 1.9
y=3
6
6
By symmetry x = 2 X (-0.833) = -1.7
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(0.833, 1.9)
( -1.7, 4)
11/10/2007
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