Real values of $k$ such that $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots

Question

There is only one real values of $k$ for which the quadratic equation $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots. Then the product of these two solutions is

What i tried.

$$kx^2+(k+3)x+k-3=0$$

for $k=0,$ i am getting $x=1$(only one integer roots)

for $\displaystyle k\neq 0, x=\frac{-k\pm \sqrt{(k+3)^2-4k(k-3)}}{2k}$

for integer roots, $(k+3)^2-4k(k-3)=p^2$

How do i solve it , Help me please

Answer 1

Recall that, for $k\not=0$, the product of the roots of the given quadratic equation is the ratio $\frac{k-3}{k}=1-\frac{3}{k}$ which in your case has to be a positive integer. This means that $k\in \{-1,-3\}$. But it is easy to check that for such values of $k$ the roots are not real. So your statement has something wrong...

On the other hand, if we ask for two integer roots then $k\in \{-1,-3,1,3\}$ and only for $k=3$ we find two integer roots, namely $0$ and $-2$.

Answer 2

There is no such $k.$

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Since the product we're looking for is $$\frac {k-3}{k},$$ first we must have $k\ne 0.$ Also, since the roots are positive integers, it follows that their product is also a positive integer. Thus, $k\ne 3$ and we also have that $k-3$ and $k$ must have the same sign.

Having noticed this, we find that the discriminant $$(k+3)^2-4k(k-3)$$ must be a nonzero square. Thus we must have $$0<(k-3)^2<12.$$ It follows that $(k-3)^2$ is either $1,4,$ or $9.$ Thus, $k-3$ must be some of $\pm 1,\pm 2, \pm 3,$ so that correspondingly, $k$ is $4,2;5,1;6,0.$

Now using the fact that $k$ is nonzero and has the same sign as $k-3$ leaves just $k$ being one of $4,5,6.$ Finally using the condition that $(k-3)/k$ be integral leaves us no such value of $k.$

Thus, the claim that there exists such a $k$ is false. Your conditions are too restrictive. Completely killed off the $k$'s.

Answer 3

Let p and q be the two positive roots, then m, the average value of the two must also be positive. That is $m = \dfrac {p + q}{2} = \dfrac {- (k + 3)}{2k} > 0$.

It should be clear that k must not be 0.

If k > 0 and after solving the above, we get k < -3. This means a contradiction.

If k < 0, a similar contradiction arises.

Answer 4

Look at it in an another way. The equation can be re-written as $$x-1+\frac{k}{3}\left(x^2+x+1\right)=0$$ which can be said to be a family of curves passing through points of intersection of the curves

$y=x-1$

and

$y=x^2+x+1$.

But then the two curves neither touch nor intersect in real points.

So no value of $k$ can create the condition you are looking for.