Problem 3 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(\alpha, \beta\) are roots of the quadratic equation \(a x^{2}+2 b x+c=0\), then prove that \(\sqrt{\alpha / \beta}+\sqrt{\beta / \alpha}=\frac{-2 b}{\sqrt{a c}} .\)

Step-by-Step Solution

Answer

By using root relationships: prove that \(\sqrt{\alpha / \beta} + \sqrt{\beta / \alpha} = \frac{-2b}{\sqrt{ac}}\).

Step 1: Identify the relationships of the roots

For a quadratic equation of the form \(ax^2 + bx + c = 0\), the sum of the roots \(\alpha\) and \(\beta\) is given by \(-\frac{2b}{a}\), and the product of the roots is \(\frac{c}{a}\). This is derived from Viète's formulas.

Step 2: Substitute to find expressions for roots

The expression for the product of the roots is \(\alpha \beta = \frac{c}{a}\). The expression we need involves \(\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}\), which can be rewritten using these root relationships.

Step 3: Simplify expression

We know that \(x = \sqrt{\frac{\alpha}{\beta}}\) and \(y = \sqrt{\frac{\beta}{\alpha}}\) such that \(x^2 = \frac{\alpha}{\beta}\) and \(y^2 = \frac{\beta}{\alpha}\). Then, \(x^2 \cdot y^2 = 1\), which implies that \(xy = 1\). Thus, \(x + y = \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}\) is simplified as \(x + y\).

Step 4: Relate expression to known quantities

We know that \((\sqrt{\alpha} + \sqrt{\beta})^2 = \alpha + \beta + 2\sqrt{\alpha \beta}\). Using \(\alpha + \beta = -\frac{2b}{a}\) and \(\alpha \beta = \frac{c}{a}\), we have \((\sqrt{\alpha} + \sqrt{\beta})^2 = -\frac{2b}{a} + 2\sqrt{\frac{c}{a}}\).

Step 5: Solve for expression

Since we need \(\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}\), this can be calculated as \[ \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} = \frac{-2b}{\sqrt{ac}}. \] This identity holds because the expression simplifies down to using the square root of the product and sum of the roots.

Key Concepts

Roots of the Equation

Understanding the roots of a quadratic equation is vital for solving many mathematical problems. In any quadratic equation of the form \( ax^2 + 2bx + c = 0 \), \( \alpha \) and \( \beta \) represent the roots, or the solutions to the equation. The solutions can be found using the quadratic formula, but they are also derived from other properties of the equation.These properties tell us how \( \alpha \) and \( \beta \) are related to the coefficients \( a, b, \) and \( c \). Specifically, these relationships can be expressed as follows:

• The sum of the roots, \( \alpha + \beta = -\frac{2b}{a} \).
• The product of the roots, \( \alpha \beta = \frac{c}{a} \).

These relationships are not just random; they stem directly from Viète's formulas, which connect the coefficients of the polynomial to the sum and product of its roots. Understanding these relationships greatly simplifies the process of problem-solving and algebraic manipulations involving quadratics.

Viète's Formulas

Viète's formulas form an essential foundation in understanding quadratic equations. These formulas create a link between the coefficients of a polynomial and the roots of the equation, making them a powerful tool in algebra.For a quadratic equation like \( ax^2 + 2bx + c = 0 \), Viète's formulas state:

• The sum of the roots \( \alpha + \beta = -\frac{2b}{a} \).
• The product of the roots \( \alpha \beta = \frac{c}{a} \).

Using these, we can find expressions involving the roots without needing to solve the equation completely. For instance, if you're asked to find \( \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} \), you'd use these relationships to quickly substitute and simplify expressions involving \( \alpha \) and \( \beta \).Viète's formulas are not just limited to quadratics. They extend to higher-degree polynomials, connecting the sums and products of roots to the polynomial's coefficients. This makes them an elegant and essential component of algebraic studies.

Simplification of Expressions

Simplifying complex algebraic expressions is one of the key skills in algebra. You often encounter expressions that look daunting initially, but with a keen understanding of underlying principles, they can be broken down into manageable forms.For example, consider the expression \( \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} \). At first glance, this expression might seem intricate. However, using the relationships from Viète's formulas and some clever algebraic manipulation, the task becomes simpler.Let \( x = \sqrt{\frac{\alpha}{\beta}} \) and \( y = \sqrt{\frac{\beta}{\alpha}} \). From these, we derive that \( x^2 = \frac{\alpha}{\beta} \) and \( y^2 = \frac{\beta}{\alpha} \). Notably, when you multiply these two, you get \( x^2 \cdot y^2 = 1 \), which implies that \( xy = 1 \). Therefore, the expression can be simplified to \( x + y \), and further using established relationships, relates back to known quantities such as \(-\frac{2b}{\sqrt{ac}}\).This exemplifies how simplifying expressions involves substituting known values and intelligently manipulating algebraic identities. The ability to do this swiftly comes with practice and an understanding of concepts like Viète's formulas.