Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to figure out how to solve the following equations:

Find the integer roots of

$$(6x^2+15xy)^{1/2}+y^2=10 \quad\text{and}\quad x+y=3$$

I tried substituting $y=3-x$ to $(6x^2+15xy)^{1/2}+y^2=10$, but I ended up having more square roots and the power of $x$ just goes higher and higher.

Please help me with a very detailed answer, as I am trying to understand how to this step by step. Thank you very very much!

share|cite|improve this question
Please click the "edit" button to see how I formatted your post. All that HTML crud you put in just wasted time because it all had to be erased. – dfeuer Oct 5 '13 at 22:57
@dfeuer thanks man. I'm a first time user – Confusedwithmath Oct 5 '13 at 23:02

Hint $y^2 \leq 10$

There are not too many integers $y$ which satisfy this inequality.....

share|cite|improve this answer
So how to do it? – Confusedwithmath Oct 5 '13 at 23:09
@Confusedwithmath Which are all the integers $y$ which satisfy $y^2 \leq 10$? – N. S. Oct 5 '13 at 23:11
1,2,3. I know that the answer is x=1 and y=2, because x+y =3 can only either be 0+3 or 1+2 or 2+1, but how do you get y^2≤10? – Confusedwithmath Oct 5 '13 at 23:19
@Confusedwithmath Those are the positive integers..... And each value of $y$ yields an $x$ such that $x+y=3$. Then you need to check if that pair satisfies the first equation... – N. S. Oct 5 '13 at 23:29
@Confusedwithmath And $y^2 \leq y^2+\sqrt{6x^2+15xy}=3$. – N. S. Oct 5 '13 at 23:30

By definition, both $\sqrt{6x^2+15xy}$ and $y^2$ are nonnegative. So the left-hand side of the equation is the sum of two nonnegative numbers, one of which is a square, whose sum is $10$. If $\lvert y \rvert \ge 4$, then $y^2 \ge 16$, and the equation can never be satisfied. Hence, you need only consider $y \in \{-3,-2,-1,0,1,2,3\}$.

Using the second equation, $x+y=3$, each of those possibilities for $y$ yields exactly one $x$, i.e. $x=3-y$. So let's start with $y=-3$. This gives $x=3-y=3-(-3)=6$. Now substituting $(x,y)=(6,-3)$ into the first equation implies \begin{align} 10 &= \sqrt{6(6)^2+15(6)(-3)} + (-3)^2 \\ &= \sqrt{-54} + 9 \\ 1 &= \sqrt{-54}, \end{align} which is obviously a contradiction.

Hope this helps!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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