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Question

ali2x2 · Answer

i saw tat question but i couldt help ;/

ali2x2 · Answer

thats cheap asto

ganeshie8 · Answer

there is really a very simple and cute solution, give it a try again :)

ali2x2 · Answer

ok :)

parthkohli · Answer

inb4 Astrophysics tags me

parthkohli · Answer

These are perfect squares. The only way we have it is that each one is zero.

ganeshie8 · Answer

Thats it!

ali2x2 · Answer

I couldnt find a way but now that i see the correct answer it clears up the fog in my mind, good job @ParthKohli :)

ganeshie8 · Answer

please finish it off @ParthKohli 
i think @praxer  is still looking for a solution

parthkohli · Answer

Oh, sorry. OS is acting up for me again. I type out things and it removes them.

anonymous · Answer

this suddenly look like ultimate troll question after you explain it lol

ganeshie8 · Answer

Haha if you don't like the simplicity, there is a complicated solution, which is quite enlightening too :)
Familiar with Cauchy-Schwarz inequality ?

parthkohli · Answer

IT REMOVED IT AGAIN!

Yes, I actually solved the same question on my test.

anonymous · Answer

no im not at that stage of mathematics yet

astrophysics · Answer

$$\sum_{i=1}^{n} (a_ix+b_i)^2 = x^2 \sum_{i=1}^{n}a^2_i+2x \sum_{i=1}^{n}a_ib_i+\sum_{i=1}^{n}b_i^2=0$$

parthkohli · Answer

The OP isn't actually too far from directly solving it. You can just expand and complete the square.

ganeshie8 · Answer

what do you mean by directly expand and complete the square ? 
i thought we will have to use cauchy-schwarz inequality

astrophysics · Answer

Yeah same, can you show it parth

astrophysics · Answer

Oh I think you mean what op was already doing

ganeshie8 · Answer

Ohkiee, I'll finish off your solution using geometry : 

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