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complete the exercises in the “Projects” section on page 331 of Mathematics in Our World.

You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example.

For Project #2, please select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects.

1. An interesting method for solving quadratic equations
came from India. The steps are
(a) Move the constant term to the right side of the
equation.
(b) Multiply each term in the equation by four times
the coefficient of the x
2
term.
(c) Square the coefficient of the original x term and
add it to both sides of the equation.
(d) Take the square root of both sides.
(e) Set the left side of the equation equal to the
positive square root of the number on the right side
and solve for x.
(f) Set the left side of the equation equal to the
negative square root of the number on the right side
of the equation and solve for x.
Example: Solve x
2
+ 3x − 10 = 0.
x
2
+ 3x = 10
4x
2
+ 12x = 40
4x
2
+ 12x + 9 = 40 + 9
4x
2
+ 12x + 9 = 49
2x + 3 = ±7
2x + 3 = 7
2x = 4
x = 2

2x + 3 = −7
2x = −10
x = −5
Try these.
(a) x
2
− 2x − 13 = 0
(b) 4x
2
− 4x + 3 = 0
(c) x
2
+ 12x − 64 = 0
(d) 2x
2
− 3x − 5 = 0




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Apr 15, 2011
Rating
starstarstarstarstar 		Completing the Square
by: Staff


The question:

complete the exercises in the “Projects” section on page 331 of Mathematics in Our World.

You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example.

For Project #2, please select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects.

1. An interesting method for solving quadratic equations came from India. The steps are:

(a) Move the constant term to the right side of the equation.

(b) Multiply each term in the equation by four times the coefficient of the x² term.

(c) Square the coefficient of the original x term and add it to both sides of the equation.

(d) Take the square root of both sides.

(e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.

(f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.

Example: Solve x² + 3x - 10 = 0.

x² + 3x = 10

4x² + 12x = 40

4x² + 12x + 9 = 40 + 9

4x² + 12x + 9 = 49

(2x + 3)² = 7²

√[(2x + 3)²] = √(7²)

2x + 3 = ±7

2x + 3 - 3 = -3 ±7

2x + 0 = -3 ±7

2x = -3 + 7 and 2x = -3 – 7

2x = 4 and 2x = -10

2x/2 = 4/2 and 2x/2 = -10/2

There are two solutions

x = 2 and x = -5


Try these.

(a) x² - 2x - 13 = 0

(b) 4x² - 4x + 3 = 0

(c) x² + 12x − 64 = 0

(d) 2x² - 3x - 5 = 0



The answer:


Try these.

(a) x² - 2x - 13 = 0

x² - 2x - 13 + 13 = 0 + 13

x² - 2x = 13

x² - 2x + 1= 13 + 1

√ (x - 1)² = ±√(14)

x - 1 = ±√(14)

x - 1 + 1 = 1 ± √(14)

x = 1 ± √(14)


(b) 4x² - 4x + 3 = 0

4x² - 4x + 3 - 3 = 0 – 3

4x² - 4x = - 3

4(x² - x) = - 3

4(x² - x)/4 = - ¾

x² - x = - ¾

x² - x + ¼ = - ¾ + ¼

x² - x + ¼ = - ½

(x - ½)² = - ½

√(x - ½)² = ±√(-½)

x - ½ = ±√(-½)

x - ½ + ½ = ½ ±√(-½)

x = ½ ±√(-½)

x = ½ ± i√(½)


(c) x² + 12x - 64 = 0

x² + 12x - 64 + 64 = 0 + 64

x² + 12x = 64

x² + 12x + 36 = 64 + 36

x² + 12x + 36 = 100

√(x + 6)² = ±√10²

x + 6 = ±10

x + 6 - 6 = -6 ±10

x = -6 + 10 and x = -6 - 10

x = 4 and x = -16


(d) 2x² - 3x - 5 = 0

2x² - 3x - 5 + 5 = 0 + 5

2x² - 3x = 5

(2x² - 3x)/2 = 5/2

x² - 3x/2 = 5/2

x² - 3x/2 + 9/16= 5/2 + 9/16

(x - ¾)² = 40/16 + 9/16

(x - ¾)² = 49/16

(x - ¾)² = (7/4)²

√(x - ¾)² = ±√(7/4)²

x - ¾ = ±(7/4)

x - ¾ + ¾ = ¾ ±(7/4)

x = ¾ ±(7/4)

x = ¾ + (7/4) and x = ¾ - (7/4)

x = 10/4 and x = - (4/4)

x = 5/2 and x = - 1




Thanks for writing.


Staff
www.solving-math-problems.com

Apr 15, 2011
Rating
starstarstarstarstar 		OMG help
by: brandy

I dont understand what it is talking about in project 2. If you can help that would be so great.
Apr 15, 2011
Rating
starstarstarstarstar 		OMG help
by: brandy

This is what it says to do. I will give you everything that it said.
Following completion of your weekly readings, complete the exercises in the “Projects” section on page 331 of Mathematics in Our World.

You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example.

For Project #2, please select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects.

The assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs.
Your introduction should include three to five sentences of general information about the topic at hand.
The body must contain a restatement of the problems and all math work, including the steps and formulas used to solve the problems.
Your conclusion must comprise a summary of the problems and the reason you selected a particular method to solve them. It would also be appropriate to include a statement as to what you learned and how you will apply the knowledge gained in this exercise to real-world situations.
Projects
1. An interesting method for solving quadratic equations
came from India. The steps are
(a) Move the constant term to the right side of the
equation.
(b) Multiply each term in the equation by four times
the coefficient of the x
2
term.
(c) Square the coefficient of the original x term and
add it to both sides of the equation.
(d) Take the square root of both sides.
(e) Set the left side of the equation equal to the
positive square root of the number on the right side
and solve for x.
(f) Set the left side of the equation equal to the
negative square root of the number on the right side
of the equation and solve for x.
Example: Solve x
2
+ 3x − 10 = 0.
x
2
+ 3x = 10
4x
2
+ 12x = 40
4x
2
+ 12x + 9 = 40 + 9
4x
2
+ 12x + 9 = 49
2x + 3 = ±7
2x + 3 = 7
2x = 4
x = 2

2x + 3 = −7
2x = −10
x = −5
Try these.
(a) x
2
− 2x − 13 = 0
(b) 4x
2
− 4x + 3 = 0
(c) x
2
+ 12x − 64 = 0
(d) 2x
2
− 3x − 5 = 0

I have email my teacher to ask him what he is wanting. I will post it if you still (like me) cant understand what he is asking. thank so much for your help.

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