solve 9x^2 + 3x - 8 = 0 by completing the square

[NEHA]

Solve the following equation by completing the square:

. . .9x^2 + 3x - 8 = 0

What I did:

. . .9/3 + 1/3x - 8/3 = 0

. . .(9/3^2 + 1/3x ) = 8/3

Is this right so far?

 

[Denis]

http://www.purplemath.com/modules/sqrquad.htm

 

[NEHA]

ok at the site....

they say

x^2 - 1/2x = 5/4

i only how they got -1/4 but how did they get 1/16
by doing this -1/4 * -1/4 ?????
so if i did the same way in my problem i would do this.... 1/9 * 1/9 * 1/9 ?????

pleaase explain

 

[stapel]

i only how they got -1/4 but how did they get 1/16

Where it said to square the -1/4, they probably squared it.

Eliz.

 

[NEHA]

thank you very much. i so understand now.

here is my work:
9x^2 + 3x - 8 = 0
9x^2 + 3x = 8
x^2 + 1/3x = 8/9
1/9 > 1/81
x^2 + 1/3x + 1/81 = 8/9 + 1/81
)x + 1/9)^2 = 73/81
x + 1/9 = +- ?(73/81) = +- ?73 / 9
x = 1/9 +- ?73 / 9
x = 1/9 + ?73/9 and x = 1/9 - ?73/9
so thats the answer

 

[tkhunny]

. . .9x^2 + 3x - 8 = 0
. . .(9/3^2 + 1/3x ) = 8/3

You will not get this until you see how this makes absolutely no sense. You must rview your fundamental properties.

1) Where did the x^2 go?
2) How did 3 turn into 1/3? or Why didn't 9 turn into 1 or 8 into 8/9?

You're just waving in the dark without these fundamental concepts.

9x^2 + 3x - 8 = 0

Divide EVERYTHING by 9

9/9 = 1
3/9 = 1/3
8/9 = 8/9

x^2 + (1/3)x - (8/9) = 0

Add 8/9

x^2 + (1/3)x = (8/9)

Calculate the secret number

(1/3) / 2 = 1/6
(1/6)^2 = 1/36

Add 1/36 to both sides

x^2 + (1/3)x + (1/36) = (8/9) + (1/36)

Factor the left side (You built it to be easy. Don't struggle with this step.)

(x + (1/6))^2 = (8/9) + (1/36)

Simplify the right side

(x + (1/6))^2 = (8/9) + (1/36) = 32/36 + 1/36 = 33/36 = 11/12

(x + (1/6))^2 = 11/12

You do the rest.

EVERY step of the way, the symbols are meaningful and we have not deviated from the originl problem. Equal signs should mean equality. Rewriting an expression should not result in an entirely different expression. Slowly, carefully...

 

[galactus]

Few know of the general 'completing the square' formula. I suppose teachers want students to learn how to do it the long way.

\(\displaystyle \L\\\underbrace{9}_{\text{a}}x^{2}+\underbrace{3}_{\text{b}}x\underbrace{-8}_{\text{c}}=0\)

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}\)

Plug in your values and you have it. Set up and all.

 

[tkhunny]

Few know of the general 'completing the square' formula.

What? Everyone learns it. It's called the "Quadratic Formula".

 

[tkhunny]

1/9 > 1/81
)x + 1/9)^2 = 73/81

You keep writing things that don't mean anything. Your notation will save you or destroy you. Make it neat and understandable.

Where did you get "1/9"?

Be careful. Write neatly.

For "square root" you can use this, "sqrt()". So, the square root of two would be sqrt(2). There are other ways to write clearly.

 

[galactus]

Few know of the general 'completing the square' formula.

What? Everyone learns it. It's called the "Quadratic Formula".

Yes. What I meant was few put it together and use a general formula for completing the square.

 

[NEHA]

x^2 + (1/6))^2 = 11/12
x^2 + 1/6^2 = 11/12
x^2 + 1/36 = 11/12
11/12 - 1/36 = 8/9
x^2 = 8/9

 

[stapel]

x^2 + (1/6))^2 = 11/12
x^2 + 1/6^2 = 11/12
x^2 + 1/36 = 11/12
11/12 - 1/36 = 8/9
x^2 = 8/9

Where is this coming from? Is the first line above a new exercise? If so, it should be in its own thread. If not, then how did you get this this point?

You were given all the steps to "x² + (1/3)x = 8/9", and "x² = 8/9" is not at all equivalent. So how did you go from what you were given, along with the step-by-step instructions and worked examples, and get to this?

Please reply with all your steps and reasoning, just as you like the tutors to give to you. Thank you.

Eliz.

 

[NEHA]

Few know of the general 'completing the square' formula. I suppose teachers want students to learn how to do it the long way.

\(\displaystyle \L\\\underbrace{9}_{\text{a}}x^{2}+\underbrace{3}_{\text{b}}x\underbrace{-8}_{\text{c}}=0\)

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}\)

Plug in your values and you have it. Set up and all.

ok wait i can do this .....way also right!