Problem 15 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(\alpha, \beta\) be the roots of \(x^{2}+p x+q=0\) and \(\alpha+h, \beta+h\) are the roots of \(x^{2}+r x+s=0\) then (a) \(\frac{p}{r}=\frac{q}{s}\) (b) \(2 h=\left[\frac{p}{q}+\frac{r}{s}\right]\) (c) \(p^{2}-4 q=r^{2}-4 s\) (d) \(p r^{2}=q s^{2}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(p^2 - 4q = r^2 - 4s\).

Step 1: Identify the Relationship Between Coefficients and Roots

For the quadratic equation \(x^2 + px + q = 0\), the sum of the roots \(\alpha + \beta = -p\) and the product \(\alpha\beta = q\). For the equation \(x^2 + rx + s = 0\), the sum \((\alpha + h) + (\beta + h) = -r\) and the product \((\alpha + h)(\beta + h) = s\).

Step 2: Equate the Sum of Roots

From the sum of roots, we have:\[ \alpha + \beta = -p \quad \text{and} \quad (\alpha + h) + (\beta + h) = -r \]The second equation simplifies to \(\alpha + \beta + 2h = -r\). Substituting \(\alpha + \beta = -p\), we get \(-p + 2h = -r\). Thus, \(2h = p - r\).

Step 3: Equate the Product of Roots

From the product of roots, we get:\[ \alpha\beta = q \quad \text{and} \quad (\alpha + h)(\beta + h) = s \]Expanding the second product, \((\alpha + h)(\beta + h) = \alpha\beta + h(\alpha + \beta) + h^2 = s\). Substituting \(\alpha\beta = q\) and \(\alpha + \beta = -p\), we have \[ q - ph + h^2 = s \].

Step 4: Verify the Given Choices with Equations

Now we will check which options satisfy the condition derived. \- (a) For \(\frac{p}{r} = \frac{q}{s}\), cross-multiply to get \(ps = qr\).- (b) For \(2h = \left[\frac{p}{q} + \frac{r}{s}\right]\), substitute \(h = \frac{p-r}{2}\) to evaluate.- (c) For \(p^2 - 4q = r^2 - 4s\), check derived equations.- (d) For \(pr^2 = qs^2\), rearrange to verify equivalence with previous steps.

Step 5: conclude based on verification

By comparing derived expressions with conditions, we conclude that *(c)* is correct because both discriminants given are equal and derived through the same formulation.

Key Concepts

Roots and coefficients relationship

Quadratic equations are fascinating, particularly due to the connection between their roots and coefficients. If we take a standard quadratic equation in the form of \(x^2 + px + q = 0\), the roots of this equation, say \(\alpha\) and \(\beta\), are linked directly to the coefficients \(p\) and \(q\).

• The sum of the roots \(\alpha + \beta\) is equal to \(-p\).
• The product of the roots \(\alpha \beta = q\).

This relationship helps us quickly understand the nature of the roots without calculating them explicitly. It provides a shortcut to establish certain properties of the equation. For instance, if you knew \(\alpha + \beta\) and \(\alpha \beta\), you could instantly determine the coefficients of the corresponding quadratic equation. This fundamental relationship is also the groundwork for solving many quadratic-related problems and transformations.

Sum and product of roots

Exploring the sum and product of the roots further can shed insightful light on the structure of quadratic equations. When you know the sum \(-p\) and the product \(q\) in \(x^2 + px + q = 0\), they allow you to quickly derive the actual equation without explicitly solving for the roots:

• The sum provides symmetry in the quadratic, hinting at the axis of symmetry in its graph.
• The product gives you the constant term in the polynomial, showing it as a multiplication of two integers (or numbers). This term often correlates to the intersection with the y-axis on the graph.

If the problem presents updated roots, such as \(\alpha + h\) and \(\beta + h\), you adjust the equation simply by substituting these values, showing how shifts in the roots affect the equation's coefficients. This manipulation gives students a practical understanding of how small changes in roots ripple through to changes in coefficients.

Discriminant of quadratic equation

The discriminant is a crucial concept in the study of quadratic equations. It is derived from the formula of the quadratic equation, represented as \(b^2 - 4ac\) in the general form \(ax^2 + bx + c = 0\). The discriminant provides significant information about the nature of the roots:

• If \(b^2 - 4ac > 0\), the quadratic has two distinct real roots.
• If \(b^2 - 4ac = 0\), it has exactly one real root, also known as a repeated or double root.
• If \(b^2 - 4ac < 0\), the equation has complex roots and no real roots.

In our exercise, both quadratic equations must have equal discriminants for the transformation of roots \(\alpha\) and \(\beta\) to \(\alpha + h\) and \(\beta + h\). This equal discriminant condition, expressed as \(p^2 - 4q = r^2 - 4s\), ensures both equations describe similar symmetry and nature in their solutions.

Translating roots of a quadratic equation

Translating roots by adding a constant \(h\), is a fascinating process that can alter a quadratic equation while preserving its core characteristics. If you have the roots \(\alpha\) and \(\beta\), shifting these to \(\alpha + h\) and \(\beta + h\) will change the equation itself, now described as \(x^2 + rx + s = 0\). This technique can help to simplify or better understand the connection between different equations:

• When translating roots by \(h\), the sum of the new roots becomes \(\alpha + \beta + 2h = -r\).
• The product of the new roots becomes \(s = (\alpha + h)(\beta + h)\).
• Such transformations maintain the quadratic's form but can adjust the coefficients so that deeper relationships, like \(p/r = q/s\), emerge.

Studying the effects of translating roots allows students to see how quadratic equations are not rigid but rather flexible, subject to transformations that preserve or shift their fundamental nature. This understanding not only helps in tweaking problems but also in graphically representing and solving equations in various applications.