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UNIT 3 : QUADRATIC
FUNCTIONS & EQUATIONS
LESSON
2: QUADRATIC EQUATIONS
Quadratic
Equations:
Review of Basic Factoring methods:
1. Common Factoring:
Factor 6x3 15x
Solution: 6x3 15x = 3x(2x2 5) **
Find the HIGHEST COMMON FACTOR for each term --- 3x
** Divide 3x into each term to get the second factor --- 2x2 5
** Check by expanding
2. Difference of Squares: Formula -- a2 b2 = (a b)(a + b)
Factor 49x2 64y2
Solution: 49x2 64y2 = (7x 8y)(7x + 8y)
Factor x2 9y2
Solution: x2 9y2 = (x 3y)(x + 3y)
3. Simple Trinomials: Form x2 + bx + c [Coefficient of x2 is 1]
Factor x2 + 5x + 6
Solution: Recall x2 + 5x + 6 = (x + __ )(x + __ ) **
We need two numbers that multiply to +6 and add to +5
** Check all pairs of factors of 6: {1, 6} adds to 7
{2, 3} adds to 5
Hence
x2 + 5x + 6 = (x + 3 )(x + 2) ** Check by expanding
4. Hard Trinomials: Form ax2 + bx + c [Coefficient of x2 does not equal 1]
Factor 6m2 5m 4
Solution:
We use the method of decomposition (although there are other methods)
We decompose the middle term -5m into two parts using the two clues:
Multiply
to (6)(-4) = -24 and
Strategy: List all pairs of factors of -24 and find the pair that adds to -5
** {1,
24} cannot obtain -5 with these two
factors; 1+24=25; 1-24= -23
** {2, 12} cannot obtain -5 with these two
factors
** {3, 8}
-8 x 3 = -24 Choose these two
factors
Hence 6m2
5m 4 = 6m2 8m + 3m 4 ** -5m broken into two parts -8m + 3m
= 2m(3m 4) + 1(3m 4) ** Group by twos and common factor
= (3m 4)(2m + 1) ** check by expanding
Zeros
[x-intercepts] of Quadratic Functions:
Example 1:
Find the
zeros (x-intercepts) of the quadratic function f(x)
= 2x2 + x 6.
Solution:
We are
trying to find the points where the graph of the function crosses the x-axis
ie the x-intercepts.
Let y =
f(x) = 0 yielding the quadratic equation
2x2 + x 6 = 0. We
solve by factoring.
2x2 + x 6 = 0
2x2 + 4x 3x 6 = 0 ** See #4 above Hard Trinomials method
of decomposition
2x(x + 2) 3(x + 2) = 0 ** Common factor by grouping first two
terms and last two terms
(x + 2)(2x 3) = 0 ** Common factor (x + 2)
Hence
either x + 2 = 0 or 2x 3 = 0
and ** Zero Product Rule if ab = 0,
then a = 0 or b = 0.
Note: Finding zeros of a quadratic function always yields a quadratic equation to solve.
Example 2:
Solve 2x2
3x + 1 = 0 by completing the square.
Solution:
Example 3:
Solve 3x2
7x + 1 = 0 by completing the square.
Solution:
Example 4:
Solve 3x2
+ 6x + 1 = 0 using the quadratic formula.
Express roots to the nearest hundredth
Solution:
Finding
the number of roots of a Quadratic Equation :
Two zeros: b2
4ac > 0 |
One zero: b2 4ac = 0 |
No zeros: b2 4ac < 0 |
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Here b2
4ac > 0 and the graph of the
quadratic function cuts the x-axis in 2 distinct points. (2 x-intercepts/zeros) The
quadratic equation has 2 real, distinct
roots. |
Here b2
4ac = 0 and the graph of the
quadratic function is tangent to the x-axis yielding 1 x-intercept/zero. The
quadratic equation has 2 real, equal
roots. Some texts say 1 real
root for this case. |
Here b2
4ac < 0 and the graph of the
quadratic function does not cut the x-axis.
The
quadratic equation has no real roots or
imaginary roots. |
Example 1:
In each
case, calculate the value of the discriminant b2 4ac and determine
the number of zeros of the quadratic function.
a) f(x) = 2x2 + 3x 1 b) g(x) = 16 x2 c) y = 3x2 2x + 5 d) y = x2 6x + 9
Solutions:
a = 2 b = 3 c= -1
a) f(x) = 2x2 + 3x 1
b2 4ac = (3)2
4(2)(-1)
= 9 + 8
= 17
Since b2
4ac > 0, there are 2 zeros.
The corresponding quadratic equation
2x2 + 3x 1 = 0 has 2
real distinct roots.
a = -1 b = 0 c = 16
b) g(x) = 16 x2
= -1x2 + 0x + 16
b2 4ac = (0)2
4(-1)(16)
= 64
Since b2
4ac > 0, there are 2 zeros.
The corresponding quadratic equation
16 x2 = 0 has 2 real
distinct roots.
a = 3 b = -2 c = 5
c) y = 3x2 2x + 5
b2
4ac = (-2)2 4(3)(5)
= 4 60
= -56
Since b2
4ac < 0, there are no zeros.
The corresponding quadratic equation
3x2 2x + 5 = 0 has no real roots.
a = 1 b = -6 c = 9
d) y = x2 6x + 9
b2 4ac = (-6)2
4(1)(9)
= 36 36
= 0
Since b2 4ac = 0, there is 1 zero. The corresponding quadratic equation x2 6x + 9 = 0 has 1 real root or real and equal roots.
Example 2:
In each case, determine the number x-intercepts/zeros of the quadratic function.
a) f(x) = 5x2 3x + 2 b) g(x) = 9x2 8.4x + 1.96 c) y = - 2x2
3x + 1 d) y = 4x2 3x + 5
Solutions:
a = 5 b = -3 c = 2
a) f(x) = 5x2 3x + 2
b2 4ac = (-3)2
4(5)(2)
= 9 40
= -31
Since b2 4ac < 0, there are no x-intercepts or zeros. The corresponding quadratic equation 5x2 3x + 2 = 0 has no real roots
a = 9 b = -4.2 c = 1.96
b) g(x) = 9x2 8.4x + 1.96
b2
4ac = (-8.4)2 4(9)(1.96)
= 70.56 70.56
= 0
Since b2 4ac = 0, there is 1 x-intercept or zero. The corresponding quadratic equation 9x2 8.4x + 1.96 = 0 has 1 real root or real and equal roots.
a = -2 b = -3 c = 1
c) y = - 2x2 3x + 1
b2 4ac = (-3)2
4(-2)(1)
= 9 + 8
= 17
Since b2
4ac > 0, there are 2 x-intercepts or
zeros. The corresponding quadratic equation - 2x2 3x + 1 = 0 has 2 real distinct roots.
a = 4 b = -3 c = 5
d) y = 4x2 3x + 5
b2 4ac = (-3)2
4(4)(5)
= 9 80
= - 71
Since b2 4ac < 0, there are no x-intercepts or zeros. The corresponding quadratic equation 5x2 3x + 2 = 0 has no real roots
Example 3:
For what
value of k does the equation kx2
4x + 2 = 0 have:
a)
exactly one root [real equal roots] ?
b) no
real roots ?
c) 2
real distinct roots ?
Solutions:
a = k b = -4 c = 2
a) For 1
real root set b2 4ac = 0
(-4)2 4(k)(2) = 0
16 8k = 0
- 8k =
-16
k =
2
b) For no real roots set b2 4ac
< 0
(-4)2 4(k)(2)
< 0
16
8k < 0
- 8k <
-16
k >
2 ** The inequality reverses when dividing
by a negative
c) For 2
real distinct roots set b2 4ac > 0
(-4)2 4(k)(2)
> 0
16 8k > 0
- 8k >
-16
k <
2 ** The inequality reverses when dividing
by a negative
Example 4:
For what
values of k will the function g(x) = x2 kx + k + 3
have exactly 1 zero?
Solution:
Exactly
1 zero means real equal roots or b2
4ac = 0.
a = 1 b = - k c = k + 3
(-k)2 4(1)(k + 3) = 0
k2
4k 12 = 0
(k 6)(k + 2)
= 0
k 6 = 0
or k + 2 = 0
k = 6 or k = -2