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

- mayankdevnani

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

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- mayankdevnani

How to write \(\large \bf 6x^2+5xy-4y^2=0\) in another form or equation?

- Mokeira

basically, just factor out

- mayankdevnani

Like i have to convert in this :-
\[\large \bf (2x-y)(3x+4y)\]
i can't judge out that how to obtain this equation ?

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## More answers

- mayankdevnani

@cwrw238 help me!!

- mayankdevnani

@SolomonZelman

- SolomonZelman

well, you just did it, no ?

- cwrw238

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

- mayankdevnani

solomon, i do it many times but i don't get this equation

- cwrw238

i draw up columns;
|dw:1408387576129:dw|

- mayankdevnani

i don't understand your method... @cwrw238

- 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

- mayankdevnani

totally confused... ??? @cwrw238

- cwrw238

yea -- it takes a bit of getting used to

- amistre64

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

- amistre64

i forgot the y :)

- cwrw238

actually therea an error in my last line in drawing it should read
4 * 2 and -1 * 3

- mayankdevnani

my BAD... i don't understand... @amistre64

- amistre64

its a method is all ...

- mayankdevnani

can you teach me step by step ??

- cwrw238

its tricky i know - it needs a lot of practice

- amistre64

it would look just like what i just typed, but sure

- mayankdevnani

but i don't know this method yet,this is first chance to see this long method...@cwrw238
hehehe

- amistre64

ax^2 + bx + c
setup the factor, assuming it has ratioanl roots
(x /a) (x /a)
and mutliply the first and last coefficients

- mayankdevnani

why its (x/a)(x/a) ?? @amistre64

- amistre64

ac = k, such that some pair of the factors of k combine to equal b

- amistre64

its not x/a there is a space left to determine the operations

- cwrw238

|dw:1408388342676:dw|

- amistre64

x ___ ( )/a

- amistre64

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

- mayankdevnani

like :-
\[\large \bf \frac{x-2}{a}\]

- mayankdevnani

this is what you mean to say

- amistre64

not quite:
at first we are just organizing the information, filling it in as we go
\[(x \pm\frac{ r_1}{a})~~(x \pm\frac{ r_2}{a})\]

- mayankdevnani

ohkk

- amistre64

if the last term is positive, then ++ or -- create a positive term
if the last term is negative, then +- or -+ create a negative term

- mayankdevnani

yeah..

- amistre64

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
\[(x +\frac{\color{red}L}{a})~~(x -\frac{ \color{red}s}{a})\]

- mayankdevnani

correct

- amistre64

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?

- mayankdevnani

another way of writing this :-
6*(-4y^2)=-24y^2
the factors being: 8y,-3y will combine to make 5y
am i correct ? @amistre64

- 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
\[(x +\frac{{4y}}{3})~~(x -\frac{{1y}}{2})\]

- mayankdevnani

correct

- amistre64

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

- amistre64

\[ax^2 + bx + c\]
\[a(x^2 + \frac{b}{a}x + \frac{c}{a})\]
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

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

- amistre64

im using the term roots incorrectly most likely .... the factors of c/a that do this and that

- myininaya

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?

- mayankdevnani

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
\[\large \bf 6x^2+8xy-3xy-4y^2=0\]
\[\large \bf 2x(3x+4y)-y(3x+4y)=0\]
\[\large \bf (2x-y)(3x+4y)=0\]
\(\large \bf \color{red}{Answer}\)^^^

- mayankdevnani

am i correct?

- mayankdevnani

@myininaya mam

- myininaya

yes you are done with the factoring

- mayankdevnani

THANK YOU GUYS for helping me......i appreciate your work..
@cwrw238 , @amistre64 (once again thank you so much)
and mostly to you @myininaya mam....

- cwrw238

yw

- amistre64

youre welcome

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