Let S be the set of all non-zero real numbers α such that the quadratic equation αx² - x + α = 0 has

Q: Wait, why do we use $D > 0$ instead of $D \ge 0$?

Great catch! The problem explicitly states that the equation must have two *distinct* real roots. If $D = 0$, the roots are real but identical (a repeated root). To satisfy the 'distinct' condition, we must strictly enforce $D > 0$.

Q: Why can we just square both sides of the inequality $\frac{\sqrt{1 - 4\alpha^2}}{|\alpha|} < 1$?

We can square both sides because both sides are guaranteed to be positive. The numerator is a square root (which is non-negative), and the denominator is an absolute value (which is positive since $\alpha \neq 0$). When both sides of an inequality are positive, squaring preserves the inequality direction.

Q: Why is the leading coefficient $a = \alpha$ so important here?

It is crucial because the formula for the difference of roots is $\frac{\sqrt{D}}{|a|}$. If $a$ were a constant, it would just be a scaling factor, but since $a$ depends on $\alpha$, it changes the magnitude of the difference as $\alpha$ varies. Always keep an eye on the leading coefficient in quadratic problems!

Question

Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^2 - x + \alpha = 0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1 - x_2| < 1$. Which of the following intervals is(are) a subset(s) of $S$?

Options:

(A) $(-1/2, -1/\sqrt{5})$

(B) $(-1/\sqrt{5}, 0)$

(C) $(0, 1/\sqrt{5})$

(D) $(1/\sqrt{5}, 1/2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Consider the quadratic equation $\alpha x^2 - x + \alpha = 0$, where $\alpha eq 0$. We seek the values of $\alpha$ such that the roots $x_1$ and $x_2$ are real, distinct, and satisfy the condition $|x_1 - x_2| < 1$. ### The Gatekeeper of Reality For the roots to be real and distinct, the discriminant $D$ must be strictly greater than zero. Given $a = \alpha$, $b = -1$, and $c = \alpha$, we calculate: $$D = (-1)^2 - 4(\alpha)(\alpha) = 1 - 4\alpha^2$$ Setting $D > 0$ yields $1 - 4\alpha^2 > 0$, which simplifies to $\alpha^2 < \frac{1}{4}$. This implies that $|\alpha| < \frac{1}{2}$. Since $\alpha eq 0$, our first constraint is: $$\alpha \in \left( -\frac{1}{2}, 0 ight) \cup \left( 0, \frac{1}{2} ight)$$ ### The Geometry of Distance To satisfy the condition $|x_1 - x_2| < 1$, we utilize the identity relating the difference of roots to the discriminant: $$|x_1 - x_2| = \frac{\sqrt{D}}{|a|}$$ Substituting our values, we obtain the inequality: $$\frac{\sqrt{1 - 4\alpha^2}}{|\alpha|} < 1$$ Since both sides are positive, we square both sides to eliminate the square root: $$\frac{1 - 4\alpha^2}{\alpha^2} < 1$$ Multiplying by $\alpha^2$ (which is positive), we get $1 - 4\alpha^2 < \alpha^2$. Rearranging the terms leads to $5\alpha^2 > 1$, or: $$\alpha^2 > \frac{1}{5} \implies |\alpha| > \frac{1}{\sqrt{5}}$$ ### The Final Convergence We must now satisfy both constraints simultaneously: $|\alpha| < \frac{1}{2}$ and $|\alpha| > \frac{1}{\sqrt{5}}$. This results in the intersection: $$\frac{1}{\sqrt{5}} < |\alpha| < \frac{1}{2}$$ The final set of values for $\alpha$ is: $$\alpha \in \left( -\frac{1}{2}, -\frac{1}{\sqrt{5}} ight) \cup \left( \frac{1}{\sqrt{5}}, \frac{1}{2} ight)$$ These intervals represent the precise range where the parabola maintains two distinct roots separated by a distance of less than one.

Select Answer:

Visualized Solution

Problem Setup

Condition for Distinct Real Roots

Calculating the Discriminant

Solving $D > 0$

Difference of Roots Formula

Setting up the Inequality

Solving the Root Inequality

Simplifying the Inequality

Second Interval for $α$

Finding the Intersection

Checking the Options

The Sigma Insight: Nature of Roots

Analyzing the Setup

The Gatekeeper of Reality

The Geometry of Distance

The Final Convergence

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