By Keith Rachels And Asef Haider

### What is a Quadratic Equation?

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.

• • • • • • •

### Solving Quadratic Functions in Standard Form

Solve x 2 + 5 x + 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 5 lots of stuff that totals zero.

x or the 6), because you can add So the first thing I have to do is factor: x 2 + 5 x + 6 = ( x + 2)( x + 3) Set this equal to zero: ( x x x + 2)( x + 3) = 0 + 2 = 0 or = –2 or x x The solution to x 2 + 3 = 0 = – 3 + 5 x + 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 say it in your head a few times. 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) : F irst - multiply the first term in each set of parenthesis: 4x * x = 4x^2 O utside - multiply the two terms on the outside: 4x * 2 = 8x I nside - multiply both of the inside terms: 6 * x = 6x L ast - 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 answer onto the remaining binomial.

### 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, ax 2 + bx + c = 0

• The quadratic formula is the equation X = -b ± b 2 – 4ac 2a If b 2 - 4ac > 0, then the equation has two real solutions If b 2 - 4ac = 0, then the equation has one real solution If b 2 - 4ac < 0, then the equation has two imaginary solutions

• Let’s find the values of x in the equation x 2 + 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 2 x = -4 ± i 100 2 x = -4 ± 10i 2 x = -2 + 5i x = -2 – 5i

### Completing the Square

• • Completing the square is a process that allows you to write an expression of the form x 2 + bx as the square of a binomial.

To complete the square of x 2 add (b ÷ 2) 2 + bx, you have to • The equation looks like this x 2 + 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 x 2 – 12x + c x 2 – 12x +( ) 2 x 2 – 12x + (-6) 2 x 2 – 12x + 36 (x - 6)(x - 6) (x – 6) 2

Now let’s solve an equation by completing the square! :D x 2 + 20x + 104 = 0 x 2 + 20x + ( ) 2 = -104 x 2 + 20x + (10) 2 = -104 + (10) 2 x 2 + 20x + 100 = -4 (x + 10) 2 = -4 (x + 10) 2 = -4 x + 10 = ± 2i x = -10 ± 2i x = -10 + 2i x = -10 – 2i

### Sum of Two Cubes

• • The sum of two cubes is represented by the equation (a 3 + b 3 ) = (a + b)(a 2 – ab + b 2 ) Let’s factor the polynomial x 3 + 27 x 3 + 27 = x 3 + 3 3 = (x + 3)(x 2 – 3x + 9)

### Difference of Two Cubes

• • • • The difference of two cubes is represented by the equation (a – b)(a 2 + ab + b 2 ) 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)(a 2 – ab + b 2 ) Difference (a – b)(a 2 + ab + b 2 )

• Let’s try to find the difference of two cubes in the equation 16u 5 – 250u 2 16u 5 – 250u 2 = 2u 2 (8u 3 – 125) Factor common monomial = 2u 2 ((2u) 3 – 5 3 ) (a 3 – b 3 ) = 2u 2 (2u – 5)(4u 2 + 10u + 25) (a - b) (a 2 + ab + b 2 )