Problem 18 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The ratio of the roots of the equation \(a x^{2}+b x+c=0\) is same as the ratio of the roots of the equation \(A x^{2}+B x+C=0 .\) If \(D_{1}\) and \(D_{2}\) are the discriminants of \(a x^{2}+b x+c=0\) and \(A x^{2}+B x+C=0\) respectively, then show that \(D_{1}: D_{2}=b^{2}: B^{2}\).

Step-by-Step Solution

Answer

Hence, we have shown that \(D_1 : D_2 = b^2 : B^2\).

Step 1: Find the roots of the given equations

Let the roots of the first equation \(a x^2 + bx + c = 0 \) be denoted by \(\alpha\) and \(\beta\), and the roots of the second equation \(A x^2 + Bx + C = 0 \) be denoted by \(\gamma\) and \(\delta\). We can find the roots using the quadratic formula: Roots of \(a x^2 + bx + c = 0 \): \[\alpha = \frac{-b + \sqrt{D_1}}{2a}\] \[\beta = \frac{-b - \sqrt{D_1}}{2a}\] Roots of \(A x^2 + Bx + C = 0 \): \[\gamma = \frac{-B + \sqrt{D_2}}{2A}\] \[\delta = \frac{-B - \sqrt{D_2}}{2A}\]

Step 2: Express the given condition in terms of the roots

We are given that the ratio of the roots of the first equation is the same as the ratio of the roots of the second equation. That is: \[\frac{\alpha}{\beta} = \frac{\gamma}{\delta}\] Now substitute the expressions for roots: \[\frac{\frac{-b + \sqrt{D_1}}{2a}}{\frac{-b - \sqrt{D_1}}{2a}} = \frac{\frac{-B + \sqrt{D_2}}{2A}}{\frac{-B - \sqrt{D_2}}{2A}}\]

Step 3: Manipulate the given condition

Now simplify and cross multiply: \[\frac{-b + \sqrt{D_1}}{-b - \sqrt{D_1}} \times \frac{-B - \sqrt{D_2}}{-B + \sqrt{D_2}} = \frac{-b - \sqrt{D_1}}{-b + \sqrt{D_1}} \times \frac{-B + \sqrt{D_2}}{-B - \sqrt{D_2}}\] Now, we use the formula \((a-b)(a+b) = a^2 - b^2\) to simplify both sides: \[\frac{b^2 - D_1}{B^2 - D_2} = 1\]

Step 4: Show the required result

From the last step, we have that: \[b^2 - D_1 = B^2 - D_2\] Now, we need to show that \(D_1 : D_2 = b^2 : B^2\). To do this, we can divide both sides of the above equation by \(D_1 D_2\): \[\frac{b^2 - D_1}{D_1} = \frac{B^2 - D_2}{D_2}\] Note that this is the same as \(\frac{D_1}{D_2} = \frac{b^2}{B^2}\), which is our required result. Hence, we have shown that \(D_1 : D_2 = b^2 : B^2\).

Key Concepts

Discriminant

The discriminant in a quadratic equation is key to understanding the nature of its roots. It is denoted by the letter \(D\) and calculated as \(D = b^2 - 4ac\) for an equation of the form \(ax^2 + bx + c = 0\). The discriminant helps us predict the nature of the roots before actually calculating them:

• If \(D > 0\), the equation has two distinct real roots.
• If \(D = 0\), there are two real and identical roots (a perfect square).
• If \(D < 0\), the equation has two complex roots.

By comparing discriminants, as we do in this exercise, you can determine how the equations' roots relate to each other in terms of these characteristics.

Root Ratio

Understanding the root ratio is essential when the problem involves comparing two quadratic equations. Suppose the equations have roots \(\alpha\), \(\beta\) for the first and \(\gamma\), \(\delta\) for the second. When we say their root ratios are the same, we express it as:
\[ \frac{\alpha}{\beta} = \frac{\gamma}{\delta} \]
It's important because matching ratios imply a proportional relationship between the roots of the equations.
This concept is useful for determining equalities without needing to know the exact values of the roots, allowing us to work within a framework of proportionality and symmetry.

Quadratic Formula

The quadratic formula is a fundamental tool for finding the roots of a quadratic equation \(ax^2 + bx + c = 0\). It is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula provides an exact calculation of the roots \(x\), incorporating the discriminant \(b^2 - 4ac\). The \(\pm\) symbol indicates that there are typically two solutions (roots): one for the plus and one for the minus, often denoted as \(\alpha\) and \(\beta\).
The quadratic formula highlights how the coefficients \(a\), \(b\), and \(c\) determine the position and ratio of the roots, encapsulating the symmetry in quadratic relationships.

Algebraic Manipulation

Algebraic manipulation allows us to simplify and solve equations by rearranging and transforming terms systematically. In this exercise, manipulating the given condition helps simplify the problem effectively. We began with expression:
\[\frac{-b + \sqrt{D_1}}{-b - \sqrt{D_1}} = \frac{-B + \sqrt{D_2}}{-B - \sqrt{D_2}}\]
Cross-multiplication and other arithmetic strategies are used while maintaining equality. The property \((a-b)(a+b) = a^2 - b^2\) is particularly useful for simplification.
These steps traverse from complex rational expressions to more straightforward comparisons. Through algebraic insight, we derive:\[\frac{b^2 - D_1}{D_1} = \frac{B^2 - D_2}{D_2}\] Effectively allowing the demonstration of proportional relationships like the original problem seeks. This highlights the power of algebra in bridging complex problems to logical solutions.