If $ax^2+bx+c=0$ has rational roots then $(a+1)x^2+bx+c=0$ cannot have rational roots

Question

I was working on quadratic equations and found the following fact that:

"If $ax^2+bx+c=0$ has rational roots then $(a+1)x^2+bx+c=0$ cannot have rational roots where $a,b,c\in\mathbb N$"

And now I have to prove or disprove it

And I know the first comment will be "show your efforts"

So I have assumed if a quadratic equation have rational roots, then $b^2-4ac$ should be a perfect square.

Applying to both equation I got $k^2-4c=l^2$ for some integer $k$ & $l$.

How to proceed now??

Please help!!

Answer 1

If I am not making a silly mistake then I think it is not true.

Suppose there are two quadratic equations $3x^2+5x+2=0$ and $2x^2+5x+2=0$.

The roots of first are $\frac{-2}{3},-1$ and roots of second are $\frac{-1}{2},-2$

Answer 2

I believe I can show that there are infinite many values of $a$, $b$ and $c$ that satisfy the conditions of your question. We know that

$$x_{1;2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ is the formula for the two solutions $x_1$and $x_2$ of the second-degree equation $ax^2+bx+c=0$. The equation has rational roots if $\Delta_1 =b^2-4ac$ is a square number.

With $(a+1)x^2+bx+c=0$ we have that $\Delta_2=b^2-4ac-4c$ which also has to be a square number (otherwise the solutions wouldn't be rational). Basically we can say that $\Delta_1$ and $\Delta_2$ are two square numbers that differ by a multiple of 4.

Note that every two number that differ by a multiple of $2$, squared will differ by a factor of $4$; for example $3^2$ and $5^2$ differ by a multiple of $4$.

With these assumptions we can construct $\Delta_1$ and $\Delta_2$ such that their difference is a multiple of $4$ and are both square numbers. We can set, as an example, $\Delta_1=49$; it follows that $\Delta_2 = 25$, they are two square numbers that differ by a multiple of $4$. Since $\Delta_1 - \Delta_2=4c$

$$\Delta_1 - \Delta_2=4c;\ 4c=49-25 \rightarrow c=6$$ to obtain the other values we use the equation

$$\Delta_1= b^2-4ac=b^2-16c=49$$ Which can be satisfied by infinite values of $a$ and $b$; if we want $b$ to be a natural number we can see that if $b^2=64$ then

$$64-40a=49 \rightarrow a=\frac{64-49}{40}=\frac58$$ In the end we have $a=\frac58$, $b=8$ and $c=6$

$$\frac58 x^2+8x+6=0 \rightarrow x_1=-\frac{4}5, x_2=-12$$ $$\left(\frac58 +1\right)x^2+8x+6=0 \rightarrow x_1=-\frac{12}{13}, x_2=-4$$

Answer 3

From $k^2 - 4 c = l^2$ you have $(k + l)(k - l) = 4 c$. This clearly admits many solutions, e.g., you can select $k - l = 2$ to get $2 l + 2 = 4 c$, or $l + 1 = 2 c$, any $c$ gives $l, b$. Other splits of factors of $4 c$ give other families of solutions.