quadratic please help a.s.a.p

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sportsaholic2397

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Alright so i have been working with quadratics over the last week or two and i have gotten to some much larger numbers and I know that there are many different solutions to solving for X when it comes to quadratics, and it has always been a problem for me to be able to figure out which solution is the right one for 3 problems i am looking at... i have done some work on all of them but I am not sure if it is right... please help and give some possible rules or tips that could help me recognize at the begining better.

Solve For X

192x^2 - 1984x + 5120 = 0.

with this one at the begining i am thinking you need to divide each side by a number, and in some problems in the past i have divided using the first number, the x^2 number. but in this case it doesnt come out to an even number, both b and c will come out as fractions, which to me seems wierd and more complicated. so i tried finding LCD or a number you can divide by all three to get an even number, which works but also causes some problems...should i change the way I start, or should I keep going?
 
Hello, sportsaholic2397!

Solve for \(\displaystyle x\!:\quad 192x^2 - 1984x + 5120 \:=\: 0\)

With this one at the begining i am thinking you need to divide each side by a number,
and in some problems in the past i have divided by the first coefficient,
but in this case it doesnt come out to an even number,
both \(\displaystyle b\) and \(\displaystyle c\) come out as fractions.
You're right: dividing is a good idea if it doesn't produce fractions.

So i tried finding the LCD. which works but also causes some problems.
This is a great idea . . . What problems did you have?
Click to expand...

\(\displaystyle \text{LCD of }192, 198, 5120\:=\:64\)

\(\displaystyle \text{Divide by 64: }\:3x^2 - 31x + 80 \:=\:0\)

. . \(\displaystyle \text{which factors: }\:(x - 5)(3x-16) \:=\:0\)

. . \(\displaystyle \text{and has roots: }\:x \;=\;5,\:\frac{16}{3}\)

 
alright thanks, i screwed up on using the proper solution. i tried using the quadratic formula. instead of getting the fctors. any advice/secrets/hints on how to better recognize which solution to use?
 
although with the factors you came up with, don't two numbers have to one add to equal a number in the original problem and multiply together to equal another?

you have (x-5)(3x-16) = 0.

-5 and -16 have to somehow equal 80 and 31? I could just not be seeing it...ha. i know it works to get 80 -5 * -16, but don't they have to equal both or maybe an explanation to help me better understand...

thank you
 
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