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

Question

If \(p x^{2}-q x+r=0\) has \(\alpha\) and \(\beta\) as its roots, evaluate \(\alpha^{3} \beta+\beta^{3} \alpha\)

Step-by-Step Solution

Answer

\(\alpha^3 \beta + \beta^3 \alpha = \frac{rq^2 - 2r^2 p}{p^3}\)

Step 1: Identify relationships among roots and coefficients

The given quadratic equation is \(px^2 - qx + r = 0\). For a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots \(\alpha + \beta = -b/a\) and the product of the roots \(\alpha \beta = c/a\). For this equation, we have \(\alpha + \beta = q/p\) and \(\alpha \beta = r/p\).

Step 2: Express \(\alpha^3\beta + \beta^3\alpha\) using known identities

We aim to express \(\alpha^3 \beta + \beta^3 \alpha\) in terms of \(\alpha + \beta\) and \(\alpha\beta\). Notice that \(\alpha^3 \beta + \beta^3 \alpha = \alpha \beta (\alpha^2 + \beta^2)\).

Step 3: Simplify \(\alpha^2 + \beta^2\)

Use the identity \((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta\). Rearranging, we have \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).

Step 4: Substitute known values

Substitute \(\alpha + \beta = q/p\) and \(\alpha\beta = r/p\) into \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).\[ \alpha^2 + \beta^2 = \left(\frac{q}{p}\right)^2 - 2 \frac{r}{p} = \frac{q^2}{p^2} - \frac{2r}{p} \]

Step 5: Calculate \(\alpha^3 \beta + \beta^3 \alpha\)

Now, substitute \(\alpha\beta = \frac{r}{p}\) and \(\alpha^2 + \beta^2 = \frac{q^2}{p^2} - \frac{2r}{p}\) back into \(\alpha^3 \beta + \beta^3 \alpha = \alpha \beta (\alpha^2 + \beta^2)\).\[ \alpha^3 \beta + \beta^3 \alpha = \left(\frac{r}{p}\right) \left(\frac{q^2}{p^2} - \frac{2r}{p}\right) \]Simplifying gives:\[ \alpha^3 \beta + \beta^3 \alpha = \frac{r}{p} \cdot \left(\frac{q^2 - 2rp}{p^2}\right) = \frac{rq^2 - 2r^2 p}{p^3} \]

Key Concepts

Roots of Quadratic Equation

When dealing with quadratic equations, one of the fundamental tasks is to determine the roots, or solutions, of the equation. The general form of a quadratic equation is given by \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are coefficients, and \(x\) represents the variable we are solving for. The roots of this equation are typically denoted as \(\alpha\) and \(\beta\). These roots can be found using the quadratic formula:

• \( \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula retrieves the roots based on the discriminant \(b^2 - 4ac\). The nature of the roots (real, repeated, or complex) depends on the value of the discriminant:

• Real and distinct roots if the discriminant is positive.
• Real and equal roots if the discriminant is zero.
• Complex roots if the discriminant is negative.

In examining any quadratic equation, comprehending the relationship between the roots and the coefficients—notably the sum \(\alpha + \beta\) and product \(\alpha \beta\)—is key for further analysis.

Vieta's Formulas

Vieta's formulas are a set of relationships between the coefficients of a polynomial and sums and products of its roots. Specifically, for a quadratic equation \(ax^2 + bx + c = 0\), they relate the roots of the equation to its coefficients:

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

These formulas simplify the process of working with polynomial equations by eliminating the need to directly solve for the roots. For instance, they allow us to express complex polynomial expressions solely in terms of the coefficients and known relationships. If one needs to evaluate expressions like \(\alpha^3 \beta + \beta^3 \alpha\), Vieta's formulas provide an efficient pathway, as demonstrated by utilizing the identities \((\alpha + \beta)^2\) or \(\alpha \beta\) to break down and calculate terms.

Polynomial Identities

Polynomial identities are powerful tools in algebra that help in simplifying complex expressions. In the context of quadratic equations, they assist in transforming or manipulating expressions involving the roots. One such identity that proves very useful is:

• \((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta\)

Using this identity, we can express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha \beta\):

• \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\)

These kinds of identities allow one to reframe and evaluate polynomial expressions without finding the roots explicitly. Thus, given the values of \(\alpha + \beta = \frac{q}{p}\) and \(\alpha\beta = \frac{r}{p}\) as coefficients of the quadratic equation, \(\alpha^3 \beta + \beta^3 \alpha\) can be rewritten and computed succinctly in terms of these known values.