      UNIT 3  : QUADRATIC FUNCTIONS & 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

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

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   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) = x2kx + 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    