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 \(\left|x_{1}-x_{2}\right|<1\). Which of the following intervals is(are) a subset(s) of \(S\) ? (A) \(\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right)\) (B) \(\left(-\frac{1}{\sqrt{5}}, 0\right)\) (C) \(\left(0, \frac{1}{\sqrt{5}}\right)\) (D) \(\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)\)

Short answer

The subsets of S are (A) \(\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right)\) and (D) \(\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)\).

Step-by-step solution

Determine the Discriminant

For the quadratic equation to have two distinct real roots, the discriminant must be greater than zero. The discriminant of the quadratic equation \(\alpha x^2 - x + \alpha=0\) is \(D = b^2 - 4ac\), where \(a = \alpha\), \(b = -1\), and \(c = \alpha\). Thus, \(D = (-1)^2 - 4(\alpha)(\alpha) = 1 - 4\alpha^2\). Set \(D > 0\) to find \(1 > 4\alpha^2\), or \(\frac{1}{4} > \alpha^2\).

Find the Interval for \(\alpha\)

Solving \(\frac{1}{4} > \alpha^2\), we find that \(\alpha\) must lie within \(-\frac{1}{2} < \alpha < \frac{1}{2}\).

Determine the Difference Between the Roots

The difference between the roots is given by \(x_1 - x_2 = \frac{\sqrt{D}}{|a|}\). We seek the subset of values for \(\alpha\) where \(|x_1 - x_2| < 1\). Substitute the discriminant to get \(\frac{\sqrt{1 - 4\alpha^2}}{|\alpha|} < 1\). Square both sides to obtain \(1 - 4\alpha^2 < \alpha^2\). This simplifies to \(5\alpha^2 > 1\) or \(\alpha^2 > \frac{1}{5}\). The solutions for \(\alpha\) are outside the interval \(-\frac{1}{\sqrt{5}} < \alpha < \frac{1}{\sqrt{5}}\).

Combine the Conditions for \(\alpha\)

Taking the intersection of the two conditions from Steps 2 and 3, we get that \(\alpha\) must satisfy both \(-\frac{1}{2} < \alpha < \frac{1}{2}\) and not be in the interval \(-\frac{1}{\sqrt{5}} < \alpha < \frac{1}{\sqrt{5}}\). The subsets of \(S\) fulfilling these conditions are \(\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right)\) and \(\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)\).

Key concepts

Real Roots of Quadratic Equations

Understanding the nature of the roots of quadratic equations is essential in algebra. A quadratic equation typically has the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(a \eq 0\). The roots of the equation are the values of \(x\) that satisfy the equation. These roots can be real or complex.

For the roots to be real, the discriminant \(D\), which is \(b^2 - 4ac\), must be greater than or equal to zero. If \(D > 0\), there are two distinct real roots, if \(D = 0\), there is a single real root (also known as a repeated or double root), and if \(D < 0\), the roots are complex.

In the exercise provided, we seek to find the set \(S\) of all non-zero real numbers \(\alpha\) that cause the quadratic equation \(\alpha x^{2} - x + \alpha = 0\) to have two distinct real roots. To satisfy this requirement, we explore the discriminant next.

Discriminant of a Quadratic Equation

The discriminant of a quadratic equation provides vital information about the nature of its roots. It is found using the formula \(D = b^2 - 4ac\) and is directly linked to the coefficients \(a\), \(b\), and \(c\) of the equation \((ax^2 + bx + c = 0\).

In our example, the quadratic equation \(\alpha x^{2} - x + \alpha = 0\) yields a discriminant of \(D = (-1)^{2} - 4(\alpha)(\alpha)\), which simplifies to \(D = 1 - 4\alpha^{2}\).

For the equation to have two distinct real roots, this discriminant must be strictly positive \(D > 0\). Solving this inequality provides us with an interval for \(\alpha\) that ensures the presence of two real roots. Within these bounds, the exercise demonstrates how to find \(\alpha\) that satisfies not just the requirement for real roots but also the additional condition regarding the difference between the roots.

Inequalities and Intervals in Quadratic Equations

In quadratic equations, inequalities come into play when we are interested in finding intervals of the coefficient \(\alpha\) that lead to desired properties of the roots, such as them being real and distinct. Intervals describe a range of values for \(\alpha\) and are essential in expressing the solutions to these inequalities.

In the exercise, once we know the interval where the discriminant is positive, it is equally important to find where the absolute difference between the roots is less than 1. To do so, we delve into the relationship between the discriminant and the roots. Specifically, \(x_1 - x_2 = \frac{\sqrt{D}}{|a|}\).

By imposing the condition \(\left|x_1 - x_2\right| < 1\), we obtain another inequality that \(\alpha\) must satisfy. The final step is to combine the two sets of conditions on \(\alpha\) to determine the specific intervals that belong to the set \(S\). Solving these inequalities gives us an understanding of the intervals where the quadratic equation meets all criteria outlined in the exercise.