Key Concepts and Formulas

• Greatest Integer Function: $[x]$ is the greatest integer less than or equal to $x$.
• Fractional Part: $\{x\} = x - [x]$, where $0 \le \{x\} < 1$.
• Factoring Quadratic Equations: $ax^2 + bx + c = 0$ can sometimes be factored into $(px+q)(rx+s) = 0$.

Step-by-Step Solution

Step 1: Rearrange the equation We want to group the terms involving $[x]$ on one side of the equation. $x^2 - 4x + [x] + 3 = x[x]$ Subtract $x[x]$ and $[x]$ from both sides: $x^2 - 4x + 3 = x[x] - [x]$ Explanation: This step prepares the equation for factorization.

Step 2: Factor both sides Factor the quadratic expression on the left and factor out $[x]$ on the right. $(x-1)(x-3) = [x](x-1)$ Explanation: Factoring simplifies the equation and allows us to identify potential solutions.

Step 3: Rewrite the equation to set it to zero Move all terms to one side of the equation. $(x-1)(x-3) - [x](x-1) = 0$ Explanation: Setting the equation to zero is a standard technique for solving equations.

Step 4: Factor out the common term $(x-1)$ Factor out the common factor of $(x-1)$ from both terms. $(x-1)((x-3) - [x]) = 0$ Explanation: This factorization is crucial for finding the solutions.

Step 5: Analyze the two possible cases For the product of two factors to be zero, at least one of the factors must be zero. This gives us two cases.

Case 1: $x-1 = 0$ $x-1 = 0$ $x = 1$ Explanation: If the first factor is zero, then the entire expression is zero, regardless of the second factor.

Step 6: Verify the solution $x=1$ Substitute $x=1$ into the original equation to check if it's a valid solution. $(1)^2 - 4(1) + [1] + 3 = (1)[1]$ $1 - 4 + 1 + 3 = 1$ $1 = 1$ Explanation: Since the equation holds true, $x=1$ is a valid solution.

Case 2: $(x-3) - [x] = 0$ $(x-3) - [x] = 0$ $x - [x] = 3$ Explanation: If the first factor is not zero, then the second factor must be zero.

Step 7: Use the definition of the fractional part Recall that $\{x\} = x - [x]$, so we can substitute this into the equation. $\{x\} = 3$ Explanation: This simplifies the equation and allows us to use the properties of the fractional part.

Step 8: Analyze the condition $\{x\} = 3$ The fractional part of any real number must satisfy $0 \le \{x\} < 1$. Explanation: This is a fundamental property of the fractional part.

Step 9: Draw a conclusion for Case 2 Since $0 \le \{x\} < 1$, the condition $\{x\} = 3$ cannot be satisfied. Therefore, there are no solutions in this case.

Step 10: Re-examine the quadratic equation approach Let $[x] = n$, where $n$ is an integer. Substituting into the original equation: $x^2 - 4x + n + 3 = nx$ $x^2 - (4+n)x + (n+3) = 0$ Using the quadratic formula: $x = \frac{(4+n) \pm \sqrt{(4+n)^2 - 4(n+3)}}{2} = \frac{(4+n) \pm \sqrt{n^2 + 4n + 4}}{2} = \frac{(4+n) \pm \sqrt{(n+2)^2}}{2} = \frac{(4+n) \pm |n+2|}{2}$

Step 11: Analyze the two sub-cases Subcase 1: $n \ge -2$. Then $x = \frac{4+n \pm (n+2)}{2}$, so $x = \frac{6+2n}{2} = 3+n$ or $x = \frac{2}{2} = 1$.

• If $x=1$, $[x]=1$, so $n=1$. This is consistent with $n \ge -2$. Thus, $x=1$ is a solution.
• If $x=3+n$, then we need $[3+n]=n$. Since $n$ is an integer, $[3+n] = 3+n$. Therefore, $3+n=n$, which gives $3=0$. This is a contradiction, so there are no solutions here.

Subcase 2: $n < -2$. Then $x = \frac{4+n \pm (-n-2)}{2}$, so $x = \frac{6+2n}{2} = 3+n$ or $x = \frac{2}{2} = 1$.

• If $x=1$, $[x]=1$, so $n=1$. This contradicts $n < -2$. So, $x=1$ is not a solution in this case.
• If $x=3+n$, then we need $[3+n]=n$. Since $n$ is an integer, $[3+n] = 3+n$. Therefore, $3+n=n$, which gives $3=0$. This is a contradiction, so there are no solutions here.

Step 12: Re-Examine Graphically

Consider the equation $(x-1)(x-3) = [x](x-1)$. If $x \ne 1$, we can divide by $(x-1)$ to get $x-3 = [x]$ or $x-[x]=3$, so $\{x\}=3$. Since $\{x\}$ must be in $[0,1)$, this is impossible. The only solution is $x=1$.

Let's check for other solutions around $x=3$. When $x \in [3, 4)$, $[x] = 3$. Our original equation becomes $x^2 - 4x + 3 + 3 = 3x \implies x^2 - 7x + 6 = 0 \implies (x-1)(x-6) = 0$. So, $x=1$ or $x=6$. But we assumed $x \in [3, 4)$, so neither of these solutions work.

Now, let's analyze a different case: When $x \in [2, 3)$, $[x] = 2$. Our original equation becomes $x^2 - 4x + 2 + 3 = 2x \implies x^2 - 6x + 5 = 0 \implies (x-1)(x-5) = 0$. So, $x=1$ or $x=5$. But we assumed $x \in [2, 3)$, so neither of these solutions work.

It seems that only one solution $x=1$ is found. However, since the problem states that there are two solutions, let us examine another case.

Let's consider $x \in [-1, 0)$, which makes $[x] = -1$. Then, the equation is $x^2 - 4x - 1 + 3 = -x \implies x^2 - 3x + 2 = 0 \implies (x-1)(x-2) = 0$, so $x=1$ or $x=2$. But we assumed $x \in [-1, 0)$, so neither solution works.

However, if the equation was instead $x^2 - 4x + [x] + 3 = x + [x]$, then $x^2 - 5x + 3 = 0$, which yields $x = \frac{5 \pm \sqrt{13}}{2}$. So, we have two solutions.

Step 13: Identify another possible solution. Let $x \in [5,6)$. Then $[x] = 5$. We have $x^2 - 4x + 5 + 3 = 5x$ $x^2 - 9x + 8 = 0$ $(x-8)(x-1) = 0$ $x=1, 8$. Neither of these are in $[5,6)$.

Let's look at some smaller numbers. When $x \in [0,1)$, $[x] = 0$. $x^2 - 4x + 3 = 0$ gives $x=1,3$.

When $x \in [1,2)$, $[x] = 1$. $x^2 - 4x + 1 + 3 = x$ $x^2 - 5x + 4 = 0$ $(x-4)(x-1) = 0$ So $x=4, 1$. $x=1$ is a solution in the interval.

When $x \in [4,5)$, $[x] = 4$. Then $x^2 - 4x + 4 + 3 = 4x \implies x^2 - 8x + 7 = 0 \implies (x-7)(x-1) = 0$. So $x=7, 1$. Neither of these are in $[4,5)$.

Step 14: Correct the algebraic mistake. The equation is $x^2 - 4x + [x] + 3 = x[x]$, which is $(x-1)(x-3) = [x](x-1)$. So $x=1$ is always a solution. What if $x \neq 1$? Then $x-3 = [x]$. So let $x = a + f$, where $a$ is an integer and $0 \le f < 1$. Then $a + f - 3 = a$. So $f=3$. But $f$ must be less than $1$. Therefore, we only have one solution.

It turns out that the only solution is $x=1$. The problem statement must have been wrong. There cannot be a second solution.

Common Mistakes & Tips

• Remember the range of the fractional part function: $0 \le \{x\} < 1$.
• When factoring, consider all possible cases, including when a factor equals zero.
• Always verify your solutions by substituting them back into the original equation.

Summary

The given equation $x^2 - 4x + [x] + 3 = x[x]$ simplifies to $(x-1)((x-3) - [x]) = 0$. Analyzing the two factors, we find that $x=1$ is a solution, and the second factor leads to the condition $\{x\} = 3$, which has no solution. Therefore, the equation has only one solution, $x=1$. However, since the problem states there are two solutions, there is likely an error in the problem statement.

Final Answer

The final answer is \boxed{1}, which does not correspond to option (A). There appears to be only one solution.