Loading presentation...

Present Remotely

Send the link below via email or IM


Present to your audience

Start remote presentation

  • Invited audience members will follow you as you navigate and present
  • People invited to a presentation do not need a Prezi account
  • This link expires 10 minutes after you close the presentation
  • A maximum of 30 users can follow your presentation
  • Learn more about this feature in our knowledge base article

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.


Make your likes visible on Facebook?

Connect your Facebook account to Prezi and let your likes appear on your timeline.
You can change this under Settings & Account at any time.

No, thanks

Quadratic Equations

Project for Math

Nate Martinez

on 11 April 2013

Comments (0)

Please log in to add your comment.

Report abuse

Transcript of Quadratic Equations

AND HOW TO SOLVE THEM QUADRATIC EQUATIONS Real-life Problem Factoring Completing the Square The square roots method Quadratic Formula There is a fire on Main Street! A fireman shoots a hose at the fire, which is 30 feet above the ground. He holds the hose at 4 feet above the ground. The water shoots out of the hose at 75 feet per second. Will the fireman be able to put out the fire? How high a fire could he put out with his hose? Factoring is best used when the numbers you are working with are relatively small and easy to factor The best time to use completing the square is when b is an even number and a is 1 The quadratic formula is a good way to solve any quadratic equation no matter what the values are What is a quadratic equation? Not a Quadratic y=4x^3+3 is not a quadratic equation because the highest power is a cube not a square. This equation would be a cubic equation. To graph a quadratic equation, you need to make sure the equation is in standard form The standard form of a quadratic is in the form y=ax^2+bx+c...let's do an example

4x^2=-2x+3 We need to add 2x to each side
4x^2+2x=+3 Now we have to subtract 3 from each side
4x^2+2x-3=0 Congratulations! You now have the standard form A quadratic equation is an equation where the highest power of a quantity is a square. Example y=2x^2 + 4x + 2 How do you graph a quadratic equation? What will 4x^2+2x-3=0 look like? Each value of a quadratic equation affects what it will look like
The "a" value affects the steepness of the parabola and whether it opens up or down
The "b" value makes the parabola move to the left/right
The "c" value is the y-intercept of the parabola Important parts of a parabola What are all those terms? Since the a-value is positive, the parabola opens up
The y-intercept of the parabola is -3 because c=-3 The vertex
The axis of symmetry
The roots Vertex...the highest/lowest point of a parabola
Axis of Symmetry...the vertical line that runs through the vertex
Roots...the points of a parabola where y=0 Make a table of data for the equation y=x^2+2x+1 x=-2
x=3 y=(-2)^2+2(-2)+1...y=4-4+1...y=1
y=(3)^2+2(3)+1...y=9+6+1...y=16 Let's put it all together now! y=x^2-2x+1 This equation is already in standard form so that's nice To find the axis of symmetry, you use x=-b/2a To find the y-value of the vertex, just plug your axis of symmetry into the equation To find the axis of symmetry we will use -b/2a or -(-2)/2(1)=2/2=1 To find the y-axis of our vertex we plug our axis of symmetry (1) into the equation
vertex=(1,0) Table of data for y=x^2-2x+1 x y x=-2
x=2 y=(-2)^2-2(-2)+1...y=4+4+1...y=9
y=(2)^2-2(2)+1...y=4-4+1...y=1 Vertex/Root The square roots method can only be used when there is no b-value An Example 4x^2-4=0 Add 4 to each side
4x^2=4 Divide each side by 4
x^2=1 Now we find the square root of each side
The roots of 4x^2-4 are positvie 1 and negitive 1 What is a square root? A square root is a number that if multiplied by itself will get a new given number. The relationship between these two numbers is that the number being multiplied by itself is the square root of the product. Examples of square roots √1=1 √16=4
√4=2 √25=5
√9=3 √36=6
√x^2=x √x^3=x√x Example when a=1 x^2+4x+3=0 The factors of 3 have to have a sum of 4
(x+3)(x+1)=0 3*1=3 and 3+1=4...Now we use the zero-product rule
x+3=0 or x+1=0 Either x+3 or x+1 equals zero because the product is 0
x={-3,-1} The roots of x^2+4x+3=0 are -3 and -1 Factoring rules:
If a=1, then the factors of c have to have a sum of b
If a>1, then the sum of wz and py has to be b
When a>1 (wx+p)(yx+z)
w and y are factors of a
p and z are factors of c Example when a>1 2x^2+7x+3=0
(2x+1)(x+3)=0 (2x*3)+(x*1)=7x so this is properly facotored
2x+1=0 or x+3=0 Zero-product rule
2x=-1 x=-3
x={-1/2,-3} The roots of 2x^2+7x+3=0 are -1/2 and -3 How to do it x^2+2x-1=0 The c value on the left side must be 0, so add 1 to both sides
x^2+2x+1=2 To make this a perfect square trinomial, divide the b value by 2 and then add its square to both sides
(x+1)^2=2 Turn the perfcet square trinomial into a squared binomial.
√(x+1)^2=±√2 Find the square root of both sides of the equations.
x+1=±√2 Subract 1 from each side and end up with...
x={-1±√2} These are the roots which is where the parabola crosses the x-axis. Example 2 x^2+6x-3=0 Add 3 to both sides
x^2+6x=3 Divide b by 2 and add its square to both sides
x^2+6x+9=3+9 Make it into a sqaure equation.
(x+3)^2=12 Take the square root of both sides.
x+3=±√12 Simpliufy the radical (The square root of 12).
x+3=±2√3 Subtract 3 from both sides.
x={-3±2√3} These are the roots. Example 3 when a≠1 4x^2+8x-8=4 Divide both sides by 4 because you need a to be 1.
x^2+2x-2=1 Add 2 to both sides.
x^2+2x=3 Divide b by 2 and add the square to both sides
x^2+2x+1=3+1 Turn it into a sqaure equation.
(x+1)^2=4 Find the square roots of both sides.
√(x+1)^2=±√4 Simplify the radical.
x+1=±2 Subract 1 from both sides.
x=-1±2 -1+2=1 and -1-2=-3
x={-3,1} The roots of 4x^2+8x-8=4 are -3 and 1. What is it? The formula for this method is -b±√b-4ac ___________ 2a How to do it: 4x^2-5x-8 5±√-5^2-4(4)(-8) ________________ Simplify what is in the radical and under the whole equations.
2(4) 5±√153 ___________ 8 Simplify the radical if possible. 5±3√17 __________ 8 Because what is in front of the radical, the first term, and the term below the equation don't have a common factor we are done. -8±√8^2-4(2)(4) Example 2: 2x+8x+4 ________________ 2(2) -8±√32 _______ 4 -8±4√2 _________ 4 x={-2±√2} Example 3: 10x^2+12x-2 -12±√12^2-4(10)(-2) _____________________ 2(10) -12±√64 _________ 20 -12±8 ______ 20 The roots are {-.2, -1} x= x= x= -12+8=-4 20 =-.2 -12-8=-20 20 =-1 x= x= x= 5^2=25 -4*4*-8=128 25+128=153
So the discriminant (what's inside the radical) is 153
If the discriminant is positive, the equation has 2 roots
If the discriminant is 0, then there is 1 root
If the discriminant is negative, then the roots are imaginary Special Cases Sum and Difference- a^2-b^2=(a+b)(a-b)
Perfect Square- a^2+2ab+b^2=(a+b)^2
The perfect square equation is also used in completing the square Example 2 2x^2=50 Divide both sides by two
x^2=25 Find the square root of both sides
x=±5 or {-5,5} 5 and -5 are our roots Example 3 x^2-2=0 Add 2 to each side
x^2=2 Find the square root of each side
x={±√2} 30 feet To solve this problem we will use the falling object module or y=-16x^2+vx+c...v can be substituted by 75, c is 4, and y is 30
Our equation is now -16x^2+75x+4=30 We need to put the equation into standard form
-16x^2+75x-26=0 or y=-16x^2+75x-26 To see if this problem is possible, we will use the previously mentioned discriminant √75^2-4(-16)(-26)
The discriminant is positive so the fireman can put out the fire! To find out how high the hose can spray, we need to figure out the y-coordinate of the vertex. The x-coordinate can be found by using -b/2a or -75/-32≈2.34...now we plug 2.34 into y=-16x^2+75x-26
The hose can put out a fire that's upto 56 high! You are making a quadratic equation into (a+b)^2 form That's All Folks by Nate Martinez and Bryan Bonnett
Full transcript