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mayankdevnani
Please Help me !! Its simple question though but i don't know how to solve it ??
How to write \(\large \bf 6x^2+5xy-4y^2=0\) in another form or equation?
basically, just factor out
Like i have to convert in this :- \[\large \bf (2x-y)(3x+4y)\] i can't judge out that how to obtain this equation ?
@cwrw238 help me!!
well, you just did it, no ?
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
solomon, i do it many times but i don't get this equation
i draw up columns; |dw:1408387576129:dw|
i don't understand your method... @cwrw238
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
totally confused... ??? @cwrw238
yea -- it takes a bit of getting used to
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) (x-s/6) 8 and 3 combine to form 5 sooo (x+8/6) (x-3/6) reduce the fractions and if there is any fraction left over, stick the bottom back in front (x+4/3) (x-1/2) (3x+4) (2x-1) thats how i tend to do it
actually therea an error in my last line in drawing it should read 4 * 2 and -1 * 3
my BAD... i don't understand... @amistre64
its a method is all ...
can you teach me step by step ??
its tricky i know - it needs a lot of practice
it would look just like what i just typed, but sure
but i don't know this method yet,this is first chance to see this long method...@cwrw238 hehehe
ax^2 + bx + c setup the factor, assuming it has ratioanl roots (x /a) (x /a) and mutliply the first and last coefficients
why its (x/a)(x/a) ?? @amistre64
ac = k, such that some pair of the factors of k combine to equal b
its not x/a there is a space left to determine the operations
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
like :- \[\large \bf \frac{x-2}{a}\]
this is what you mean to say
not quite: at first we are just organizing the information, filling it in as we go \[(x \pm\frac{ r_1}{a})~~(x \pm\frac{ r_2}{a})\]
if the last term is positive, then ++ or -- create a positive term if the last term is negative, then +- or -+ create a negative term
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 \[(x +\frac{\color{red}L}{a})~~(x -\frac{ \color{red}s}{a})\]
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?
another way of writing this :- 6*(-4y^2)=-24y^2 the factors being: 8y,-3y will combine to make 5y am i correct ? @amistre64
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 \[(x +\frac{{4y}}{3})~~(x -\frac{{1y}}{2})\]
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
\[ax^2 + bx + c\] \[a(x^2 + \frac{b}{a}x + \frac{c}{a})\] then finding the roots such that the factors of c/a add up to b/a gets us \[a(x+r_1)(x+r_2)\]
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
im using the term roots incorrectly most likely .... the factors of c/a that do this and that
Another example: 3x^2-7xy -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?
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 \[\large \bf 6x^2+8xy-3xy-4y^2=0\] \[\large \bf 2x(3x+4y)-y(3x+4y)=0\] \[\large \bf (2x-y)(3x+4y)=0\] \(\large \bf \color{red}{Answer}\)^^^
yes you are done with the factoring
THANK YOU GUYS for helping me......i appreciate your work.. @cwrw238 , @amistre64 (once again thank you so much) and mostly to you @myininaya mam....