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mayankdevnani
Group Title
Please Help me !!
Its simple question though but i don't know how to solve it ??
 one month ago
 one month ago
mayankdevnani Group Title
Please Help me !! Its simple question though but i don't know how to solve it ??
 one month ago
 one month ago

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mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
How to write \(\large \bf 6x^2+5xy4y^2=0\) in another form or equation?
 one month ago

Mokeira Group TitleBest ResponseYou've already chosen the best response.0
basically, just factor out
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
Like i have to convert in this : \[\large \bf (2xy)(3x+4y)\] i can't judge out that how to obtain this equation ?
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
@cwrw238 help me!!
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
@SolomonZelman
 one month ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.0
well, you just did it, no ?
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
well this is the way i do these  you need to look at factors of the first coefficient and the last that is 6 and 4 6 could be 6 * 1 or 2 * 3 and 4 could be 4*1 or 4 * 1 , or 2 * 2 now we have to find a way to combine these factor so as to get the middle coefficient +5
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
solomon, i do it many times but i don't get this equation
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
i draw up columns; dw:1408387576129:dw
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
i don't understand your method... @cwrw238
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
so 4 links with 2 and 1 with +3 4 and 2 on the outside and 1 and 3 on inside giving (2x  y)( 3x + 4y)  this works for me
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
totally confused... ??? @cwrw238
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
yea  it takes a bit of getting used to
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
6x^2 +5xy 4y^2=0 assume it factors rationally, multiply the first and last coeffs, and set up the solution to divide off the first term afterwards 6*4 = 24, factors of 24 that combine to 5 c since the last term is negative, we have different signs, since the middle term is positive, the largest factor is postivie (x+L/6) (xs/6) 8 and 3 combine to form 5 sooo (x+8/6) (x3/6) reduce the fractions and if there is any fraction left over, stick the bottom back in front (x+4/3) (x1/2) (3x+4) (2x1) thats how i tend to do it
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i forgot the y :)
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
actually therea an error in my last line in drawing it should read 4 * 2 and 1 * 3
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
my BAD... i don't understand... @amistre64
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
its a method is all ...
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
can you teach me step by step ??
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
its tricky i know  it needs a lot of practice
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
it would look just like what i just typed, but sure
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
but i don't know this method yet,this is first chance to see this long method...@cwrw238 hehehe
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
ax^2 + bx + c setup the factor, assuming it has ratioanl roots (x /a) (x /a) and mutliply the first and last coefficients
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
why its (x/a)(x/a) ?? @amistre64
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
ac = k, such that some pair of the factors of k combine to equal b
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
its not x/a there is a space left to determine the operations
 one month ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
dw:1408388342676:dw
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
x ___ ( )/a
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im not sure why it works out like that, but i recall when i went thr the class that the hundreds of problems we did it just sort of patterned out this way
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
like : \[\large \bf \frac{x2}{a}\]
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
this is what you mean to say
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
not quite: at first we are just organizing the information, filling it in as we go \[(x \pm\frac{ r_1}{a})~~(x \pm\frac{ r_2}{a})\]
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
ohkk
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
if the last term is positive, then ++ or  create a positive term if the last term is negative, then + or + create a negative term
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
yeah..
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
since the last term in negative, lets go with: \[(x \color{red}+\frac{ r_1}{a})~~(x \color{red}\frac{ r_2}{a})\] now, when we add up to get the middle term, the sign tells us which is larger, in this case +5 means the the positive part is larger than the negative part \[(x +\frac{\color{red}L}{a})~~(x \frac{ \color{red}s}{a})\]
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
correct
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
now, we can determine L and s in a more usual way: 6 * 4y^2 = 24y^2 the factors being: 8y,3y will combine to make 5y agreed?
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
another way of writing this : 6*(4y^2)=24y^2 the factors being: 8y,3y will combine to make 5y am i correct ? @amistre64
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
yes \[(x +\frac{\color{red}{8y}}{6})~~(x \frac{ \color{red}{3y}}{6})\] then we simply reduce the fractions, and if any are left oever, stick the bottom back in front \[(x +\frac{{4y}}{3})~~(x \frac{{1y}}{2})\]
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
correct
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
ideally, we would have factored out a 6 to start with and ended up with:\[6~~(x +\frac{{4y}}{3})~~(x \frac{{1y}}{2})\] \[3(x +\frac{{4y}}{3})~~2(x \frac{{1y}}{2})\] \[(3x +4y)~~(2x y)\] which is just visually putting the bottom of the fraction back in front where we got it from to start with
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[ax^2 + bx + c\] \[a(x^2 + \frac{b}{a}x + \frac{c}{a})\] then finding the roots such that the factors of c/a add up to b/a gets us \[a(x+r_1)(x+r_2)\]
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
To factor ax^2+bx+c you find two factors of a*c that have product ac and sum b. It is the same way if you have something in this form: ax^2+bxy+cy^2 Find two factors of a*c that have product ac and sum b. Say b=m+n and ac=mn Then ax^2+mxy+nxy+cy^2 Now we factor by grouping... So pretend we have x^2+6xy+9y^2 ac=9 and b=6 ac=9=3(3) and b=3+3 so we have x^2+3xy+3xy+9y^2 Then factor by grouping x(x+3y)+3y(x+3y) (x+3y)(x+3y) (x+3y)^2
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im using the term roots incorrectly most likely .... the factors of c/a that do this and that
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Another example: 3x^27xy 6y^2 Let's see if you can factor this one... what is a,b, and c? then you must find a*c. then you must find two factors of a*c such that when multiplied together you get ac and when you add them you get b. Let's do this slowly.. First question what is a,b, and c?
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
Sorry :( for not answering your question,but i know how to do it : Here is method(suggest by you): ab=24 and a+b=5 Now,we get a=8 and b=3 So,we get \[\large \bf 6x^2+8xy3xy4y^2=0\] \[\large \bf 2x(3x+4y)y(3x+4y)=0\] \[\large \bf (2xy)(3x+4y)=0\] \(\large \bf \color{red}{Answer}\)^^^
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
am i correct?
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
@myininaya mam
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
yes you are done with the factoring
 one month ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.3
THANK YOU GUYS for helping me......i appreciate your work.. @cwrw238 , @amistre64 (once again thank you so much) and mostly to you @myininaya mam....
 one month ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
youre welcome
 one month ago
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