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Math Help... Don't know how to solve it

YaBaBo

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Jan 27, 2010
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2
(2x^2-8x-42)/6x^2=(x^2-9)/x^2-3x

How do you solve this problem? Please list the steps.. Thank You :D
 

Loren

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Aug 28, 2007
Messages
1,299
Before I spend any more time on this do you mean (2x^2-8x-42)/6x^2=(x^2-9)/(x^2-3x)?<<<Note the added parenthesis.

Your entry >>> (2x^2-8x-42)/6x^2=(x^2-9)/x^2-3x means \(\displaystyle \frac{2x^2 - 8x - 42}{6x^2}=\frac{x^2 - 9}{x^2} - 3x\). Is that what you mean?
 

YaBaBo

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Jan 27, 2010
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2
the x^2 and -3x is together... so its x^2-3x
 

mmm4444bot

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Oct 6, 2005
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10,158
YaBaBo said:
(2x^2 - 8x - 42)/6x^2 = (x^2 - 9)/(x^2 - 3x) Thank you for clarifying this.

How do you solve this problem? Carefully! (There's a common, fatal mistake just waiting to happen.) :)

Please list the steps. Okay. See below.

Step 1: Factor the numerator and denominator in each of the two algebraic ratios

(Hint: You do not need to completely factorize 2x^2 - 8x - 42. It won't hurt anything, if you do, but you'll end up multiplying the binomials back together later, I'm thinking. So, you can get by with simply factoring out a 2.)

Step 2: Cancel common factors, to simplify each algebraic ratio

Step 3: Subtract the ratio on the righthand side from both sides of the equation

Step 4: Get a common denominator, and combine the two ratios into one ratio

Step 5: Multiply both sides of the resulting equation by the denominator, in order to clear the fraction

Step 6: Expand and combine like-terms on the lefthand side

Step 7: Solve the resulting quadratic equation, which turns out to be the following

2x^2 + 13x + 21 = 0

Step 8: Check both solutions, by substituting them into the original proportion (one at a time) and simplifying both sides to ensure that you end up with true statements

If you want more help, please show whatever work that you can or explain what you're thinking, so that we might determine why you're stuck.

Cheers ~ Mark
 

syedrox123454

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Jan 28, 2010
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4
its not 2x^2-8x-42)/6x^2=(x^2-9)/x^2-3x


its [(2x^2-8x-42)/6x^2]/[(x^2-9)/x^2-3x]

you divide it there not equal
- syed
 

mmm4444bot

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syedrox123454 said:
its [(2x^2-8x-42)/6x^2]/[(x^2-9)/x^2-3x]

you divide it there not equal

- syed
Hi Syed:

How do you know that the equals sign should be a fraction bar?

Also, you changed (x^2 - 9)/(x^2 - 3x) back to (x^2 - 9)/x^2 - 3x.

This contradicts YoBaBo's correction !

Cheers ~ Mark
 

syedrox123454

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Jan 28, 2010
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4
I found the same question in my book and there was a divide sign not a equal sign
 

courtneycoles27

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Feb 10, 2010
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1
this is a simple problem can any body answer this what is square root of 1764
 

mmm4444bot

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courtneycoles27 said:
this is a simple problem can any body answer this what is square root of 1764
Courtney, you posted your question at the end of somebody else's discussion.

In the future, please use the [NEWTOPIC] button on the board's index page, to start your own discussion.

My calculator says that 42^2 = 1764.

Cheers ~ Mark
 
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