factorise 10root2 x^2 + 3x - 9root2

\[10\sqrt{2} x ^{2}+3x-9\sqrt{2}\]

\[10\sqrt{2}x^2+3x-9\sqrt{2}\]?

yes pls...

ok i would not want to so this by had, so lets cheat

how???

set = 0 and solve using quadratic formula. get the zeros of \[r_1,r_2\] and then write \[(x-r_1)(x-r_2)\]

the zeros are \[\frac{3\sqrt{2}}{5}\] and \[-\frac{3}{2\sqrt{2}}\]

i get (-3+3root(89))/20root2 and (-3-3root(89))/20root2 as the roots... :/

today i am too tired and brain-dead and did not feel like doing it as it baffled me a litttle so i put it out for u brainiacs to do it for me ... o( :{)

wow...i'll have to check my math. mine are not pretty but satellite's are

so we can factor as \[(5x-3\sqrt{2})(2\sqrt{2}x+3)\]

i cheated. found the zeros, then factored. that is probably easiest way in this case. otherwise i am stuck

you can check by multiplication that this works. think you can even do it in your head.

first is right. last is right

Guys, the student who brought this to me has to solve it by "splitting the middle term" method as he has not been taught any other method yet......

well then lets split it.

All DETAILED help appreciated today !!!!!

$%^&* really?

what is the gimmick? we want to split the middle term. guess maybe we can work backwards and see what it should be

see what i meant... its a foxy one !!!

3x=15x-12x try that

my hero!

still not clear how it works though. can you elaborate?

yeah that is the crux i cud not work out finding factors of 180 was no problem further steps eluded me

now i am intrigued. can you explain?

although i have to admit i like my way better!

i think i got it......

(sqrt(2))^4=4

\[10\sqrt{2}x ^{2} + 15x - 12x -9\sqrt{2}\]

\[10\sqrt{2}x^2+15x-12x-9\sqrt{2}=5x(2\sqrt{2}x+3)-3(4x+3\sqrt{2})=5x(2\sqrt{2}x+3)-3((\sqrt{2})^4x+3\]

\[5x(2\sqrt{2}x +3) -3\sqrt{2}(2\sqrt{2}x + 3)\]

\[5x(2\sqrt{2}x+3)-3((\sqrt{2})^4x+3\sqrt{2})=5x(2\sqrt{2}x+3)-3\sqrt{2}(2\sqrt{2}x+3)\]

yes! and the rest is easy

\[(5x-3\sqrt{2})(2\sqrt{2}x + 3)\]

@Harkiat. I was almost there...i went the other way and pulled the 2root(2)x + 3 out from each :)

pls check my working if its okay???

thats what i got above

gj

I get the same answer...not sure i'd be able to motivate the split without some trouble though.

@myininaya thanks for the good work !! ALL OTHERS - Thanks for getting my juices flowing again. Combined "brain-storming" defeats the PROBLEM once more !!!!!

lorda mercy i am sticking to finding the zeros, or at the very least completing the square. god work though!

i mean good work. maybe god's work too, who knows?

i like this is a good bonus question for peeps in college algebra

sure if you want them to really hate you!

lol they wouldn't get it

JAI HO !!! ( as in the song from SLUMDOG MILLIONAIRE)

MEDALS to ALL - every thread of contribution appreciated, with special word of praise for myininaya (Is that yr real name???) !!!!!!

no it isn't i want to call my daughter myininaya

i made it up im in love with the name

what does it mean???

i guess i could make up a meaning lol

where r u from, I am from Delhi - India

@ satellite CONGRATS FOR CROSSING 1000 mark !!!!

united states

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