let [x] denotes the greatest integer in x then in the interval [0,3] the numver of solutions of the equation x²-3x+[x] = 0 is

To determine the number of solutions for the equation $x^2-3x+[x]=0$ in the interval $[0,3]$, where $[x]$ denotes the greatest integer less than or equal to $x$, we need to analyze the equation by considering the behavior of the greatest integer function within the specified interval.

Understanding the Greatest Integer Function

The greatest integer function, $[x]$, takes a real number $x$ and rounds it down to the nearest integer. In the interval $[0,3]$, the values of $[x]$ will be:

• For $0\le x<1$, $[x]=0$
• For $1\le x<2$, $[x]=1$
• For $2\le x<3$, $[x]=2$
• At $x=3$, $[x]=3$

Breaking Down the Equation

We can rewrite the original equation $x^2-3x+[x]=0$ for each segment of the interval based on the value of $[x]$. This gives us three separate cases to analyze:

Case 1: $0\le x<1$ (where $[x]=0$)

The equation simplifies to:

$x^2-3x+0=0$ or $x^2-3x=0$

Factoring gives:

$x(x-3)=0$

The solutions are $x=0$ and $x=3$. However, only $x=0$ is valid in this interval.

Case 2: $1\le x<2$ (where $[x]=1$

The equation becomes:

$x^2-3x+1=0$

Using the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have:

$x=\frac{3\pm \sqrt{(-3{)}^2-4\cdot 1\cdot 1}}{2\cdot 1}=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}$

This gives us two solutions:

$x_1=\frac{3+\sqrt{5}}{2}\approx 2.618$ and $x_2=\frac{3-\sqrt{5}}{2}\approx 0.382$. Only $x_1$ is valid in this interval.

Case 3: $2\le x<3$ (where $[x]=2$

The equation now is:

$x^2-3x+2=0$

Factoring gives:

$(x-1)(x-2)=0$

The solutions are $x=1$ and $x=2$. However, only $x=2$ is valid in this interval.

Counting the Solutions

Now, let's summarize the valid solutions from each case:

• From Case 1: $x=0$
• From Case 2: $x=\frac{3+\sqrt{5}}{2}$ (approximately 2.618)
• From Case 3: $x=2$

Thus, the total number of solutions in the interval $[0,3]$ is three: $0$, $2$, and approximately $2.618$.

In conclusion, the equation $x^2-3x+[x]=0$ has a total of three solutions in the interval $[0,3]$.