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

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

Mokeira
 one year ago
Best ResponseYou've already chosen the best response.0basically, just factor out

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

mayankdevnani
 one year ago
Best ResponseYou've already chosen the best response.3@cwrw238 help me!!

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

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.0well 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

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

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

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

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.0so 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

mayankdevnani
 one year ago
Best ResponseYou've already chosen the best response.3totally confused... ??? @cwrw238

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.06x^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

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0its a method is all ...

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

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

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

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

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

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

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0im 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

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

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

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0since 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 \[(x +\frac{\color{red}L}{a})~~(x \frac{ \color{red}s}{a})\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0now, 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?

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0ideally, 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

amistre64
 one year ago
Best 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 \[a(x+r_1)(x+r_2)\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2To 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

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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Another 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?

mayankdevnani
 one year ago
Best ResponseYou've already chosen the best response.3Sorry :( 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 \[\large \bf 6x^2+8xy3xy4y^2=0\] \[\large \bf 2x(3x+4y)y(3x+4y)=0\] \[\large \bf (2xy)(3x+4y)=0\] \(\large \bf \color{red}{Answer}\)^^^

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2yes you are done with the factoring

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