Finding $k$ for 2 Non-Negative Roots of $x^2-2x\lfloor x \rfloor +x-k=0$

• MHB
• anemone
In summary, the relationship between the two roots of the equation x^2-2x&#8943;x +x-k=0 is that they are both non-negative and their sum is equal to k. To determine the value of k, you can use the quadratic formula and plug in the values of the roots for x. There are restrictions on the values of the roots, as they must be non-negative for the equation to be valid. This equation cannot have more than two non-negative roots, as it is a quadratic equation. The floor function is used to round down any decimal values of x to the nearest integer, ensuring that the roots are non-negative.
anemone
Gold Member
MHB
POTW Director
Find all values of $k$ for which the equation $x^2-2x\lfloor x \rfloor +x-k=0$ has two distinct non-negative roots.

Note that for any real number $x$ we always have

$\lfloor x \rfloor \le x < \lfloor x \rfloor +1$

Then putting $\lfloor x \rfloor =y$ and $x-\lfloor x \rfloor=z$ we have

$z^2+z-y^2+y-k=0$, where $y$ is an integer and $z\in [0,\,1)$.

Expressing $z$ in terms of $y$ yields

$z=\dfrac{-1\pm\sqrt{1+4(y^2-y+k}}{2}$.

Since $z\ge 0$, we have

$z=\dfrac{-1+\sqrt{1+4(y^2-y+k)}}{2}$

So $0\le \dfrac{-1+\sqrt{1+4(y^2-y+k)}}{2}<1$, or equivalently,

$0\le y^2-y+k<2$

If $x_1>x_2$ are two distinct non-negative roots of the given equation, then $y_1>y_2$. Indeed, since $\lfloor x_i \rfloor=y_i$ and $x_i-\lfloor x_i \rfloor =z_i$ $(i=1,\,2)$, we have $y_1\ge y_2$. Assume that $y_1=y_2$. In this case, by $z=\dfrac{-1+\sqrt{1+4(y^2-y+k}}{2}$, $z_1=z_2$ and so $x_1=x_2$. This is impossible.

Thus $y_1>y_2$. From $0\le y^2-y+k<2$, it follows that

$|y_1^2-y_1-y_2^2+y_2|<2$, or equivalently,

$(y_1-y_2)|y_1+y_2-1|<2$.

Note that $y_1,\,y_2$ are integers, and so $y_1-y_2\ge 1$. Then the last inequality shows that $|y_1+y_2-1|=0,\,1$.

For $|y_1+y_2-1|=0$: $y_1+y_2=1$ and hence $y_1=1,\,y_2=0$.
For $|y_1+y_2-1|=1$: $y_1+y_2=2$ and hence $y_1=2,\,y_2=0$. But these values do not satisfy $(y_1-y_2)|y_1+y_2-1|<2$.

Thus we see that if the given equation has two non-negative distinct roots $x_1>x_2$, then $\lfloor x_1 \rfloor=1,\,\lfloor x_2 \rfloor=0$. Hence,

$x_1=\dfrac{\sqrt{1+4k}+1}{2},\\x_2=\dfrac{\sqrt{1+4k}-1}{2}$

Obviously, this equation cannot have more than two distinct roots.

Finally, it follows that the possible range of $k$ is $0\le k <2$.

Nice!

-Dan

1. What is the significance of finding k for 2 non-negative roots of x^2-2x[x]+x-k=0?

The equation x^2-2x[x]+x-k=0 represents a quadratic equation with two variables, x and k. By finding the value of k, we can determine the specific values of x that satisfy the equation and thus, the two non-negative roots of the equation. This can help us solve real-world problems involving quadratic equations.

2. How do you find the value of k for 2 non-negative roots of x^2-2x[x]+x-k=0?

To find the value of k, we can use the quadratic formula: k = x^2-2x[x]+x. This formula will give us the value of k for any given value of x. We can also use graphical methods or algebraic manipulation to find the value of k.

3. Can the value of k be negative for 2 non-negative roots of x^2-2x[x]+x-k=0?

No, the value of k cannot be negative for 2 non-negative roots. This is because the equation x^2-2x[x]+x-k=0 has two non-negative roots, which means that both roots must be greater than or equal to 0. Therefore, the value of k must be equal to or greater than 0 to satisfy the equation.

4. How does the value of k affect the non-negative roots of x^2-2x[x]+x-k=0?

The value of k directly affects the non-negative roots of the equation. If k is too large, it may result in complex roots or no real roots at all. If k is too small, it may result in negative roots. Therefore, finding the appropriate value of k is crucial in determining the non-negative roots of the equation.

5. Are there any specific techniques for finding the value of k for 2 non-negative roots of x^2-2x[x]+x-k=0?

Yes, there are specific techniques for finding the value of k. Some common techniques include using the quadratic formula, factoring, and completing the square. Additionally, we can also use trial and error or approximation methods to find the value of k.

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