Quadratics

standard form- completing the square

Learning Goals

I am able too....


  • Complete the square = the purpose of completing the square is too figure out the vertex form of a quadratic relation
example:


complete the square for this equation 4x^2 +40x-300=0


simplify the equation by diving each term by 4 so it now becomes = x^2+ 10x - 75 = 0


since this equation cannot be factored we will solve this and make it into a perfect square.


now we have to divide the 'b' in this equation by 2 and square it, which in this case becomes 25


x^2+10x +25 -25= 75

we now move the -25 to the right side

x^2+10x+25=75+25

the final equation would be :


(x+5)^2=100

  • know the quadratic formula
x=-b+*sqaure root* b^2-4ac

2a


The quadratic formula is just another to way to solve for the x-intercetps of an equation if factoring is not possible for that particular relationship. This way you an simply plug in the numbers and depending on the discriminant, you could get more then one x intercepts.


example:


3x+6x-429=0

so our a= 3 b=6 c=-429 so know we put all these numbers into the equation.


x= -6 + *square root* 6-4(3)(429)

2(3)

x= -6 + *square root* 6+5148

6

x= -6 + *square root*6+72

6

x= 6+ 72 OR x= -6-72

6 6

x= 12 x=-11

  • Understand how a discriminant affects the number of solutions

if D >0= 2 solutions


If D<0= no solution


if D=0= 1 solution

Summary of unit

This unit covers quadratics in standard form which is ax^2+bx+c. It revolves around learning how to find the vertex by completing the square. As well as using the quadratic formula to find the x- intercepts of a relationship which will help graph the parabola.

Word problem

A model rocket is launched from the deck in Niall's backyard and the path followed by the rocket can be modelled by the relation h= -5t^2+100 t + 15


a) What is the initial height of the rocket?

-> the initial height of the rocket is 15.

b) What is the heigh of the rocket after 2s?

h= -5(2)^2+ 100 (2)+ 15

h= -20+ 200+15

h=195

There for the height of the rocket after 2s was 195.

c) What is the maximum height of this rocket?

t= b/29

t= 100/2(-5)

t= -100/-1

t-10

h=-5(10)^2 +100 (10)+15

h= -500+ 1000+15

h= 515

There for the maximum height for this rocket was 515m .

d) How long did the rocket take to reach this height?

-> the rocket took 10s to reach this height.

e) How long was the rocket above 200 m, to the nearest second?

h= -5t^2 +100t+15

200= -5t^2 + 100t+15

0= -5t^2+100t+15- 200

0= -5t^2+100t+185

x= -100+*square root* 100^2 - 4(-5)(-185)/10

x= -100+*square root* 10000-3700/10

x= -100+ *square root* 79.4

x= -100+79.4

x= -2.06

AND

x= -100-79.4

x= 17.94

-2.06+ 17.94

= 15.88

= 16

There the rocket was in the air above 200 for 16s.