imqwerty
  • imqwerty
fun question :D
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
imqwerty
  • imqwerty
How many integers 'x' are there such that \[x^2 - 3x -19 \] is divisible by 289? :)
anonymous
  • anonymous
28
imqwerty
  • imqwerty
nd hw did u get that :)

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More answers

imqwerty
  • imqwerty
thats wrng btw
anonymous
  • anonymous
ok
imqwerty
  • imqwerty
no thats not correct :)
imqwerty
  • imqwerty
don't guess :)
sohailiftikhar
  • sohailiftikhar
so do u knoe correct answer huh ? ok give me options if u can
imqwerty
  • imqwerty
yes i knw the answer :D nd this is a subjective question
sohailiftikhar
  • sohailiftikhar
can u give me options ?
imqwerty
  • imqwerty
ok your options are - a)5 b)9 c)289 d)none of these
sohailiftikhar
  • sohailiftikhar
u are asking about numbers that are divisible by 289 or only a certain number for x?
imqwerty
  • imqwerty
m asking that for how many values of x the expression \[x^2 -3x -19 \] is divisible by 289
sohailiftikhar
  • sohailiftikhar
no one
imqwerty
  • imqwerty
:) can u give any reason to support your answer?
sohailiftikhar
  • sohailiftikhar
yes because only 289 is divisible by 289 not any other number ...u can't get 289 by putting any value of x
sohailiftikhar
  • sohailiftikhar
ganes is typing a reply ...:)
ganeshie8
  • ganeshie8
Suppose \( x^2-3x-19\equiv 0\pmod{289}~~\color{red}{\star}\) Since \(289=17^2\), it must necessarily be the case that \(x^2-3x-19 \equiv 0\pmod{17}\) multiply \(4\) through out, complete the square and simplifying gives : \[(2x-3)^2\equiv 0\pmod{17}\] so the solution is \(x\equiv 10\pmod{17}\)
ganeshie8
  • ganeshie8
plugging that solution in \(\color{red}{\star}\) : \(x^2-3x-19\equiv (10+17k)^2-3(10+17k)-19 \\ \equiv 100+340k+289k^2-30-51k-19\\ \equiv 51+289k+289k^2\\ \equiv 51\\ \not\equiv 0\pmod{289} \) contradiction \(\blacksquare\)
imqwerty
  • imqwerty
yes :D
ganeshie8
  • ganeshie8
im pretty sure there are other simpler ways to solve this but what i showed you uses the standard method to solve any quadratic congruence... no tricks!
imqwerty
  • imqwerty
yes :) there is another simple method to solve it.
ganeshie8
  • ganeshie8
il let others try :)
imqwerty
  • imqwerty
ok :)
ParthKohli
  • ParthKohli
\[x^2 - 3x - 19 = 289t\]\[\Rightarrow x^2 - 3x - 19 - 289t = 0\]This equation has integral roots. So for that to happen, \(D\) is a perfect square.\[\Rightarrow 9 + 4(19 - 289t) = 85 - 4\cdot 289 t = 17 (5 - 68t)\]is a perfect square. The second factor must be divisible by 17 to make it a perfect square. (And that is only one condition.) But that's not possible, right?
ganeshie8
  • ganeshie8
That's pretty clever!
ParthKohli
  • ParthKohli
In general, for integral roots of the equation \(x^2 + bx + c = 0\), the discriminant must be a perfect square. Proof:\[ x= \frac{-b\pm \sqrt{b^2 - 4c}}{2}\]Now if \(b\) is even then \(D\) is also even, hence its root which is an integer is also even. So we've shown that the roots will be even. Similarly, we can think about the odd-case.
ParthKohli
  • ParthKohli
@Astrophysics I love discriminants.
ganeshie8
  • ganeshie8
I like your method more! the discriminant must be a perfect square for the roots to be rational, otherwise we get a nonsimplifying radical in the quadratic formula...
ParthKohli
  • ParthKohli
No, it actually means that the roots are not only rational, but integral. :P
imqwerty
  • imqwerty
clever and neat @ParthKohli
ganeshie8
  • ganeshie8
how do you know the top is always even ?
ParthKohli
  • ParthKohli
I wrote the proof above. Take it casewise. First, take b to be even, then odd. If we know that D is a perfect square, then the top will be even. Surprising.
ganeshie8
  • ganeshie8
Ahh okay, the only rational roots of \(x^2+bx+c\) are integers by rational root theorem
ParthKohli
  • ParthKohli
Oh, that's a good insight. Thanks. Why in the world didn't I think about that... >_<
imqwerty
  • imqwerty
my method - \[x^2 -3x -19\]\[(x-10)(x+7)+51\] suppose 289 divides \[x^2 - 3x -19\] so for some integer x. then 17 divides it ...we knw that 17 divides 51 nd also 17 must divide (x-10)(x+7). Since 17 is prime it must divide (x-10) or (x+7). nd we see that (x+7-(x-10) = 17 so when 17 divides one of (x-10) or (x+7) it must also divide the other :) so this means 17^2=289 can divide (x-10)(x+7) it follows that 289 can divide 51 :D which is impossible . So there is no integer x for which 289 divides \[x^2 - 3x -19\]
ganeshie8
  • ganeshie8
Actually i was considering the case \(ax^2+bx+c\) when \(a\ne 1\) somehow and thought ur earlier statement must be wrong
ganeshie8
  • ganeshie8
Wow that is a cool one too! now we have 3 different methods
imqwerty
  • imqwerty
:)
ParthKohli
  • ParthKohli
Hmm, it seems like you guys got the same contradiction in the end. Cool answers.
ganeshie8
  • ganeshie8
yes first and third methods look more or less similar, but the 3rd solution looks better because it avoids all that congruence mess
ganeshie8
  • ganeshie8
2nd solution is my fav so far though
ParthKohli
  • ParthKohli
Thanks, @geniusie8.

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