2
$\begingroup$

The graphs $y=3(x−h)^2+j$ and $y=2(x−h)^2+k$ have y-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer x-intercepts. Find $h$.

The answer is an integer between 1 and 999.

I substituted $0$ for $x$ in both equations and I was able to derive that $2016=3k-2j.$ I am not really sure where to go from here.

$\endgroup$
  • 1
    $\begingroup$ I'm curious whether this is a typo $\color{blue}{3}(x-h)\color{red}{2} + j$ $\endgroup$ – caverac May 30 '18 at 23:34
  • 1
    $\begingroup$ This red $2$ is certainly an exponent. We have two parabolas. $\endgroup$ – Piquito May 30 '18 at 23:36
  • 1
    $\begingroup$ Probably should be $3(x - h)^2 + j$ $\endgroup$ – Phil H May 30 '18 at 23:37
  • 1
    $\begingroup$ In that case $3k - 2j = 2016$ $\endgroup$ – Phil H May 30 '18 at 23:42
  • $\begingroup$ oops i'm bad at latex and at adding apparently $\endgroup$ – Dude156 May 30 '18 at 23:51
2
$\begingroup$

We are given that there are two positive integers $x$ such that $3(x-h)^2 + j =0$. For one of those values of $x$, let $p = x-h$; then we must have $j = -3p^2$.

Similarly, we must have $k = -2q^2$ for some integer $q$.

Now, when we set $x=0$ to find the $y$-intercept, we have $$3(0-h)^2 - 3p^2 = 2013, \qquad 2(0-h)^2 - 2q^2 = 2014$$ or $h^2-p^2 = 671$ and $h^2-q^2 = 1007$.

These both factor as differences of squares: $(h+p)(h-p)=671$ and $(h+q)(h-q)=1007$. Coincidentally, both $671$ and $1007$ have very few factorizations: $671 = 671 \cdot 1 = 61 \cdot 11$ and $1007 = 1007 \cdot 1 = 53 \cdot 19$.

Setting $h+p = 61$, $h-p = 11$, $h+q = 53$, $h-q = 19$ turns out to be the only alternative which works. It gives us $h = 36$, $p = 25$ (which means $j=-1875$), and $q=17$ (which means $k = -578$).

$\endgroup$
  • $\begingroup$ Thanks for the cool solution! That was pretty smart! $\endgroup$ – Dude156 May 31 '18 at 0:29
1
$\begingroup$

Here is a hint which should let you finish off the problem: Note that each parabola has positive integer x-intercepts. So for some positive integer n, $$3(n-h)^2+j=0\implies j = -3(n-h)^2$$ So j is a square multiplied by -3. Similarly, k is a square multiplied by -2. So let $$j=-3a^2, k=-2b^2$$ Now $$3k-2j = 2016 \implies 6a^2 - 6b^2 = 2016 \implies (a+b)(a-b) = 336$$ You should now consider factors of 336 to find some values a,b and then test which ones work using $$3h^2 + j = 2013, 2h^2 + k = 2014$$ Hope this helps.

$\endgroup$
  • $\begingroup$ That was pretty ingenious! Thanks for the help! $\endgroup$ – Dude156 May 31 '18 at 0:14
  • $\begingroup$ Btw, how did you come to thought of substituting a instead of that n-h stuffs? $\endgroup$ – Dude156 May 31 '18 at 0:15
  • $\begingroup$ You don't have to substitute a, and you can leave it as n-h. However, for the sake of simplicity, since n is unknown and arbitrary, I thought it was better to substitute it for a (and the other for b). I'm glad that I was able to help! $\endgroup$ – Ethan May 31 '18 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.