Multiple Methods for Multiple Methods for Solving QuadraticsSolving Quadratics

Section P.5Section P.5

Definition: Quadratic Equation in x

A quadratic equation in x is one that can be writtenin the form

ax + bx + c = 02

where a, b, and c are real numbers with a = 0.

Now, our Now, our fivefive methods of solving for today… methods of solving for today…

Method #1: FactoringMethod #1: Factoring

1. Set the equation equal to zero1. Set the equation equal to zero

2. Completely factor the quadratic equation2. Completely factor the quadratic equation

3. Set each factored part separately equal to zero,3. Set each factored part separately equal to zero, solve for the unknown in eachsolve for the unknown in each

Guided Practice: Guided Practice: Solve for x: x – 6x + 5 = 02

Factored form: (x – 5)(x – 1) = 0

x – 5 = 0 OR x – 1 = 0

Zero Factor PropertyZero Factor Propertyx = 5 OR x = 1

Guided Practice:Guided Practice:

Solve by factoring: x – 5x = 142 x = –2, 7x = –2, 7

Guided Practice: Guided Practice:

Solve by factoring: 6x – 7x – 24 = 02 x =x = –– 3322

,, 8833

Hint: use the “grouping method”

Method #2: GraphicallyMethod #2: Graphically

1. Graph the quadratic equation (set an appropriate1. Graph the quadratic equation (set an appropriate viewing window)viewing window)

2. Calculate the zeros (x-intercepts) using your 2. Calculate the zeros (x-intercepts) using your graphergrapher

Back to the first problem…Back to the first problem…

Solve for x: x – 6x + 5 = 02

This time, use your grapher!!!

3. Note: if you use your graph, you must always 3. Note: if you use your graph, you must always include a sketch of that graph with your solution! include a sketch of that graph with your solution!

Method #3: Extracting Square RootsMethod #3: Extracting Square Roots

1. Get the “squared” term by itself on one side of1. Get the “squared” term by itself on one side of the equationthe equation

2. Take the square root of both sides (remember to2. Take the square root of both sides (remember to take either a take either a positivepositive or or negativenegative answer when answer when “ “extracting” the roots!)extracting” the roots!)

3. Solve for the unknown (two separate equations)3. Solve for the unknown (two separate equations)

Guided Practice:Guided Practice:

Solve by extracting square roots: (2x – 1) = 92

Solution: x = 2, –1Solution: x = 2, –1

Guided Practice:Guided Practice:

Solve by extracting square roots: 6w – 13 = 15 – 3w2 2

w =w = ––2 72 7

33++Solution:Solution:

Method #4: Completing the SquareMethod #4: Completing the Square

1. Collect the “x” terms by themselves on one side1. Collect the “x” terms by themselves on one side of the equationof the equation

2. Factor the “x” terms so that the x coefficient is 12. Factor the “x” terms so that the x coefficient is 1

3. Add (b/2) to both sides of the equation3. Add (b/2) to both sides of the equation

22

22

4. Factor the new “x” terms in the equation4. Factor the new “x” terms in the equation

5. Solve for x by extracting roots, as in the previous5. Solve for x by extracting roots, as in the previous methodmethod

Guided Practice:Guided Practice:

Solve by completing the square: 4x – 20x + 17 = 02

52

2x Solution:

So, what is this So, what is this “most famous “most famous formula?”…formula?”…

(method 5 by the way…)

I’m speaking, of course, of the

Quadratic Formula:Quadratic Formula:The solutions of the quadratic equation ax + bx + c = 0,where a = 0, are given by the quadratic formula

2

x =– b b – 4ac

2a

+ 2

(note: this formula is derived via the “completing(note: this formula is derived via the “completingthe square” method…)the square” method…)

Guided Practice:

Solve for x (using the quad formula): 3x – 6x = 52

x = 1 +2 6

3= 2.633, –0.633

Can we support this answer graphically?

Solution:

Guided Practice:Solve by using the quadratic formula:

22 3 1 0x x 1

1,2

x Solution:Solution:

Other Types of Problems…Solve the equation graphically and with a table:

x = 1.325x = 1.325

x – x – 1 = 03

Not an Not an exact exact answer answer rounded to the thousandth rounded to the thousandth

Other Types of Problems…Solve the equation graphically(this may be done 2 different ways graphically):

x = –1.627, 2.827x = –1.627, 2.827

5x – 6x – 23 = 02

Solution:

Remember to show your graph as part of your solution!

Other Types of Problems…Solve the equation algebraically (support graphically):

x = 3.5, –2.5x = 3.5, –2.5|2x – 1| = 6

Graphical Support???Graphical Support???

Solution:Solution:

Other Types of Problems…Solve the equation graphically (think intersections of 2 separate equations):

x = –3, 1x = –3, 1

x = |2x – 3|2

Solution:

Other Types of Problems…A particular rugby pitch is 30 meters longer than it is wide,and the area of the pitch is 8800 m . What are the dimensionsof this particular pitch?

80 meters x 110 meters80 meters x 110 meters

2

Area: (length)(width) = 8800

30 8800w w

80w

But here, length = width + 30…

2 30 8800 0w w

Solution:

Whiteboard Problems: Solve by using the quadratic formula (not the program):

2 5 3x x

1.532,3.264x Solution:

Whiteboard Problems: Solve the equation graphically:

No Solution!No Solution!

|3x – 2| = x – 11

2

Why??Why??

Whiteboard Problems:Whiteboard Problems:

Solve by graphing: 3x + 2x – 9 = 02

x = 1.431, –2.097x = 1.431, –2.097

How could we get exact answers???

Quadratic equation (not the program)!

Solution:

Whiteboard Problems:Whiteboard Problems:

Solve by completing the square:

3x – 6x – 7 =02

4 46

3 3x Solution:

Whiteboard problems…Solve the equation graphically:

x = –1.942, 0.558, 1.384x = –1.942, 0.558, 1.384

x = x –31

3

1

2

Solution:

Homework: p. 50-51 1-23 odd, 31, 43

• Quiz tomorrow on sections 3, 4, and 5 !!!

• Remember I’m here before school if you need help on some material!