Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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 ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@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
i forgot the y :)
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
x ___ ( )/a
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 \[\large \bf 6x^2+8xy-3xy-4y^2=0\] \[\large \bf 2x(3x+4y)-y(3x+4y)=0\] \[\large \bf (2x-y)(3x+4y)=0\] \(\large \bf \color{red}{Answer}\)^^^
am i correct?
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....
youre welcome

Not the answer you are looking for?

Search for more explanations.

Ask your own question