Problem 4 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(\alpha\) and \(\beta\) are roots of the quadratic equation \(x^{2}-p x+36=0\) and \(\alpha^{2}+\beta^{2}=9\), then find the value of \(p\).

Step-by-Step Solution

Answer

The value of \(p\) is 9 or -9.

Step 1: Use Vieta's formulas

Given the quadratic equation \(x^2 - px + 36 = 0\), by Vieta's formulas, we know \(\alpha + \beta = p\) and \(\alpha \beta = 36\). These expressions come from the standard form \(ax^2 + bx + c = 0\), where sum of roots \(\alpha + \beta = -b/a\) and product of roots \(\alpha \beta = c/a\).

Step 2: Express \(\alpha^2 + \beta^2\) in terms of \(\alpha + \beta\) and \(\alpha\beta\)

Recall the identity: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\). Substitute \(\alpha + \beta = p\) and \(\alpha\beta = 36\) into the identity: \((\alpha + \beta)^2 - 2\alpha \beta = p^2 - 2(36) = p^2 - 72\).

Step 3: Set up the equation from the given condition

We are given that \(\alpha^2 + \beta^2 = 9\). Therefore, set \(p^2 - 72 = 9\).

Step 4: Solve for \(p^2\)

From the equation \(p^2 - 72 = 9\), solve for \(p^2\) by adding 72 to both sides: \(p^2 = 9 + 72 = 81\).

Step 5: Solve for \(p\)

Since \(p^2 = 81\), take the square root of both sides: \(p = \pm 9\). Therefore, \(p\) can be either 9 or -9.

Key Concepts

Vieta's Formulas

Vieta's formulas are a wonderful shortcut in the world of quadratic equations. They link the coefficients of the equation directly with the sum and product of its roots, providing an easier way to understand relationships without fully solving the equation. For a quadratic of the form \(ax^2 + bx + c = 0\):

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

In our problem, since the quadratic equation is \(x^2 - px + 36 = 0\), Vieta's gives us:

• Sum of the roots: \(\alpha + \beta = p\).
• Product of the roots: \(\alpha \beta = 36\).

This connection helps us move further in establishing relationships such as \(\alpha^2 + \beta^2\) using the values of \(p\) and \(36\). It's an essential component in approaching these types of questions.

Sum and Product of Roots

Understanding the sum and product of roots in quadratic equations lays the groundwork for deeper analysis, like finding unknown coefficients. When we talk about the sum and product, it's all about how these two properties link to the coefficients of the polynomial.Let's explore this further:

• The sum of roots \((\alpha + \beta)\) equals the coefficient from the \(x\) term (preceded by a negative sign in standard quadratic form), divided by the coefficient from the \(x^2\) term.


• The product of the roots \((\alpha \beta)\) equals the constant term, \(c\), divided by the leading coefficient, \(a\).

By using these formulas:
Given \(\alpha + \beta = p\) and \(\alpha \beta = 36\), you can derive expressions such as \(\alpha^2 + \beta^2\), using polynomial identities. This can further lead to manipulating these expressions to solve for unknowns like \(p\).

Polynomial Identities

Polynomial identities are like trusty sidekicks when deciphering the behavior of expressions that contain roots. They save the day by providing shortcuts. One of the key identities to remember when dealing with sum of squares is:

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

This identity lets you convert the expression for the squares of roots into something manageable using what you already know: the sum and product of roots.
For our quadratic equation, we used it in the context of deriving:

• \(\alpha^2 + \beta^2 = p^2 - 72\)

By reducing these expressions to simpler forms with known variables, we solve edges of polynomial puzzles without solving explicitly for each root. This makes polynomial identities not just theoretical tools, but practical guides in unpacking polynomial mysteries effectively.