```Solving Quadratic Equations
By Keith Rachels
And Asef Haider
A quadratic function has the form
y= ax^2 + bx + c
where “a” does not equal to 0.
The quadratic form shown above is written
in standard form. In the following slides, there
are two other forms of the quadratic function.
Vertex Form
• The vertex form of a quadratic function is
shown as :
y= a(x-h)^2 + k
- The vertex is (h,k) and the axis of
symmetry is x=h
Intercept Form
*The quadratic formula written in intercept
form looks like this:
y= a(x-p)(x-q)
• In this form, the x-intercepts are p and q.
Standard Form
• Solve x2 + 5x + 6 = 0.
• This equation is in standard form. The quadratic must first be
factored, because it is only when you MULTIPLY and get zero
that you can say anything about the factors and solutions. You
can't conclude anything about the individual terms of the
unfactored quadratic (like the 5x or the 6), because you can add
lots of stuff that totals zero.
• So the first thing I have to do is factor:
• x2 + 5x + 6 = (x + 2)(x + 3)
• Set this equal to zero:
• (x + 2)(x + 3) = 0
x + 2 = 0 or x + 3 = 0
x = –2 or x = – 3
• The solution to x2 + 5x + 6 = 0 is x = –3, –2
F.O.I.L
•
•
FOIL means first, outside, inside, last. That's not too hard to remember if you
You use FOIL to multiply the terms inside the parenthesis in a specific order:
first, outside, inside, last. Here's how to solve (4x + 6)(x + 2):
First - multiply the first term in each set of parenthesis: 4x * x = 4x^2
•
Outside - multiply the two terms on the outside: 4x * 2 = 8x
•
Inside - multiply both of the inside terms: 6 * x = 6x
•
Last - multiply the last term in each set of parenthesis: 6 * 2 = 12
•
Now just add everything together to get 4x^2 + 14x + 12. This method only
works easily with two binomials. To multiply something complicated like (4x +
6)(5x - 3)(15 - x), just do FOIL on two of the binomials and then distribute the
•
Example using FOIL
•
•
•
•
Example:
Multiply the following: (2x-5)(x-4)
Solution:
Just follow the letters in FOIL:
First: 2x*x=2x^2.
Outside: 2x*(-4)=-8x.
Inside: -5*x=-5x.
Last: (-5)*(-4)=20.
• Sum it all up and you get: (2x^2-13x+20).
The quadratic formula is an equation that helps
you find the roots of an equation you can’t
normally factor.
Before you apply an equation to the quadratic
formula, you must write the equation in
standard form, ax2 + bx + c = 0
• The quadratic formula is the equation
X = -b ± b2 – 4ac
2a
If b2 - 4ac > 0, then the equation has two real solutions
If b2 - 4ac = 0, then the equation has one real solution
If b2 - 4ac < 0, then the equation has two imaginary
solutions
• Let’s find the values of x in the equation
x2 + 4x + 29
We see that a = 1 b = 4 and c = 29
x = -(4)± (4) 2 – 4(1)(29)
2(1)
x = -4 ± 16 – 116
x = -4 ± 10i
2
2
x = -2 + 5i
x = -4 ± i 100
x = -2 – 5i
2
Completing the Square
• Completing the square is a process that allows
you to write an expression of the form x2 + bx
as the square of a binomial.
• To complete the square of x2 + bx, you have to
• The equation looks like this
x2 + bx + (b÷2) 2 = (x + (b÷2)) 2
• Let’s find the value of c that makes the
expression a perfect square trinomial and
then write the equation as the square of the
binomial! :D
• x2 – 12x + c
x2 – 12x +( ) 2
x2 – 12x + (-6) 2
x2 – 12x + 36
(x - 6)(x - 6)
(x – 6) 2
Now let’s solve an equation by completing the
square! :D
x2 + 20x + 104 = 0
x2 + 20x + (
) 2 = -104
x2 + 20x + (10) 2 = -104 + (10) 2
x2 + 20x + 100 = -4
x + 10 = ± 2i
(x + 10) 2 = -4
x = -10 ± 2i
x = -10 + 2i
(x + 10) 2 = -4
x = -10 – 2i
Sum of Two Cubes
• The sum of two cubes is represented by the
equation (a3 + b3 ) = (a + b)(a2 – ab + b2 )
• Let’s factor the polynomial x3 + 27
x3 + 27 = x3 + 33
= (x + 3)(x2 – 3x + 9)
Difference of Two Cubes
• The difference of two cubes is represented by
the equation (a – b)(a2 + ab + b2 )
• It looks similar to the sum of two cubes but
the positive and negative signs a swapped
between the b in the first set of parenthesis
and the ab in the second set of parenthesis
• Sum (a + b)(a2 – ab + b2 )
• Difference (a – b)(a2 + ab + b2 )
• Let’s try to find the difference of two cubes in
the equation 16u5 – 250u2
Factor common
16u5 – 250u2 = 2u2(8u3 – 125)
monomial
= 2u2((2u)3 – 53)
(a3 – b3)
= 2u2(2u – 5)(4u2 + 10u + 25)
(a - b)
(a2 + ab + b2)
```