If sqrt(2x²) - sqrt(3x) + k = 0 has 2 solutions, which are sin(phi) and cos(phi) (0
yes, @Loser66
Well, I don't know where to start this to be honest, Lol.
It is a problem from monbukagakusho, a scholarship program. It's supposed to be algebra.
I'll try. But I don't have the answers because I took the examination today.
Yes, it was.
I do not want to know it for fun, I wanna learn how to solve it :P
sqrt(2x²) - sqrt(3x) + k = 0 sqrt(2x²) + k = -sqrt(3x) square both sides and get a quadratic in x
oh, sin (phi) and cos (phi) are roots, there fore. sin(phi)*cos (phi) = c/a
once you have the quadratic, since you're given roots = \(\sin \phi , \cos \phi\) , use sum of roots and product of roots, you will get another quadratic in k, you can solve k.
yep. we meet there. hehe
Oh well, I didn't think of that... :P
@rsadhvika how can you write the long sentence and have latex at the middle? all the time It jumps to next line when I use latex.
there are multiple ways to do this, that quadratic method is the most elegant i can think of another way wud be :- since you're given two roots, plug them directly in the given radical equation - play with identities and cancel out \(\phi\) terms, you will end up with some equation in k
use this @Loser66 ``` \( latex code here \) ```
:) latex is fun isn't it :P
I tried plugging in directly, but it gets really tiring and the examination is pretty long, to the point we have little time. Thanks you both, going to try it here. Medal goes to who? :o
I don't want medal
give it to rsadhvika
lol if i were that girl and want to hide, i wouldnt tell the truth anyways :)
Thanks, guys. I have another question but it is in calculus part.
ask him , post it, he will answer you
i had a look at that - cursory glance, i gelt that question is more of geometry o.o
*felt
I'm asked to find its critical point, though.
yeah calculus scares meh lets see
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