© All Rights reserved @ LearnWithDash
Step 1: Understand the Problem
The given equation is
$-\,3\bigl(x - [x]\bigr)^2 + 2\bigl(x - [x]\bigr) + a^2 = 0,$
where $[\,x\,]$ denotes the greatest integer less than or equal to $x$, often called the integral part of $x$, and $\{x\} = x - [x]$ is the fractional part of $x$. We need to find the values of the real parameter $a$ for which this equation has no integral solution.
Step 2: Rewrite the Equation in Terms of Fractional Part
If $x$ is an integer, then $\{x\} = 0$. However, the problem specifies that the equation must have no integral solution, which effectively means we consider only cases where $\{x\} \neq 0$ (so $x$ is not an integer). Set $\{x\} = y$ for simplicity. Then $0 \le y < 1$ (the usual range for the fractional part). The equation becomes:
$-\,3y^2 + 2y + a^2 = 0.$
Step 3: Solve for the Fractional Part $y$
We solve the quadratic in $y$:
$-3y^2 + 2y + a^2 = 0.$
Rewrite as $3y^2 - 2y - a^2 = 0.$
The solutions for $y$ from the quadratic formula are
$y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-a^2)}}{2 \cdot 3}
= \frac{2 \pm \sqrt{4 + 12a^2}}{6}.$
So
$y = \frac{2 + \sqrt{4 + 12a^2}}{6}$ or $y = \frac{2 - \sqrt{4 + 12a^2}}{6}.$
Step 4: Impose the Fractional Part Constraint $0 \le y < 1$
Since $y = \{x\}$ is the fractional part, it must satisfy $0 \le y < 1$. We now analyze each root with this constraint. We also have the condition that the equation should have no integral solution, which means $y$ should not be forced to be 0.
Case 1: $y = \frac{2 + \sqrt{4 + 12a^2}}{6}$
We must check if this value lies in $[0,1)$. Starting with the inequality
$0 \le \frac{2 + \sqrt{4 + 12a^2}}{6} < 1.$
The left inequality $0 \le \frac{2 + \sqrt{4 + 12a^2}}{6}$ is always true because $\sqrt{4 + 12a^2}$ is non-negative and 2 is positive. Thus, $2 + \sqrt{4 + 12a^2} \ge 0.$
The important inequality is
$\frac{2 + \sqrt{4 + 12a^2}}{6} < 1
\implies 2 + \sqrt{4 + 12a^2} < 6
\implies \sqrt{4 + 12a^2} < 4
\implies 4 + 12a^2 < 16
\implies 12a^2 < 12
\implies a^2 < 1
\implies -1 < a < 1.$
But if $a=0$, then the original equation reduces to $-3y^2 + 2y = 0$, or $y(2 - 3y) = 0$. This gives $y=0$ or $y=\frac{2}{3}$. The $y=0$ case would make $x$ integral, contradicting the requirement that there is no integral solution. Hence $a=0$ must be excluded. This leaves the open intervals $(-1,0)$ and $(0,1)$ as possibilities from this case.
Case 2: $y = \frac{2 - \sqrt{4 + 12a^2}}{6}$
Check $0 \le \frac{2 - \sqrt{4 + 12a^2}}{6} < 1.$
For the expression to be non-negative, $2 - \sqrt{4 + 12a^2} \ge 0 \implies \sqrt{4 + 12a^2} \le 2 \implies 4 + 12a^2 \le 4 \implies 12a^2 \le 0 \implies a^2 = 0 \implies a=0.$
As explained above, $a=0$ leads to a contradiction with the requirement that $y \neq 0$ if we want no integral solution. Thus, this case does not give a valid range for $a$ apart from $a=0$, which is disallowed.
Combining these observations, the only valid range for $a$ that prevents $x$ from being an integer solution is when $-1 < a < 0$ or $0 < a < 1.$
Step 5: Final Range for $a$
Therefore, all possible values of $a$ for which the equation has no integral solution lie in the interval:
$(-1,\,0)\,\cup\,(0,\,1).$
