Problem 142 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of ' \(a\) ' for which one root of the quadratic equation \(\left(a^{2}-5 a+3\right) x^{2}+(3 a-1) x+2=0\) is twice as large as the other, is (A) \(\frac{2}{3}\) (B) \(-\frac{2}{3}\) (C) \(\frac{1}{3}\) (D) \(-\frac{1}{3}\)

Step-by-Step Solution

Answer

The value of \(a\) is \(\frac{2}{3}\).

Step 1: Assign Variables for Roots

Let the roots of the quadratic equation be \( r \) and \( 2r \), where one root is twice the other.

Step 2: Use Vieta's Formulas

By Vieta's formulas, the sum of the roots \( r + 2r \) is equal to \(-\frac{b}{a}\) of the quadratic equation \((a^2-5a+3)x^2+(3a-1)x+2=0\). So, \( 3r = -\frac{3a-1}{a^2-5a+3} \).

Step 3: Use Product of Roots

By Vieta's formulas, the product of the roots \( r \times 2r = 2r^2 \) is equal to \( \frac{c}{a} \). Hence, \( 2r^2 = \frac{2}{a^2-5a+3} \).

Step 4: Solve for \(r\) from Product Equation

From step 3, we have \( r^2 = \frac{1}{a^2-5a+3} \). Thus, \( r = \frac{1}{\sqrt{a^2-5a+3}} \).

Step 5: Substitute \(r\) in Sum Equation

From step 2, substitute \( r = \frac{1}{\sqrt{a^2-5a+3}} \) into \( 3r = -\frac{3a-1}{a^2-5a+3} \), giving \( 3\left( \frac{1}{\sqrt{a^2-5a+3}} \right) = -\frac{3a-1}{a^2-5a+3} \).

Step 6: Equate and Solve for \(a\)

Multiply both sides by \( a^2-5a+3 \) and simplify to get \( 3\sqrt{a^2-5a+3} = -(3a-1) \). Solve this equation for \( a \).

Step 7: Simplify and Find Roots

Squaring both sides of equation for simplification, we have \( 9(a^2-5a+3) = (3a-1)^2 \). Simplify to get the expression, \( 9a^2 - 45a + 27 = 9a^2 - 6a + 1 \).

Step 8: Further Simplification

Cancel \( 9a^2 \) on both sides and rearrange the terms: \(-45a + 27 = -6a + 1\). Thus, \(-39a = -26\).

Step 9: Solve for \(a\)

Solving \(-39a = -26\) gives \( a = \frac{2}{3} \).

Key Concepts

Vieta's Formulas

Vieta's formulas are a set of equations relating the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form \(ax^2 + bx + c = 0\), these formulas state:

• The sum of the roots \(r_1 + r_2 = -\frac{b}{a}\).
• The product of the roots \(r_1 \cdot r_2 = \frac{c}{a}\).

Using Vieta's formulas makes it easier to analyze a quadratic equation without solving for the roots directly. They relate the roots to the equation's coefficients, allowing for a deeper understanding of quadratic equations.
In our problem, we used Vieta's formulas to express both the sum and the product of the roots in terms of \(a\). This approach helps to determine the value of \(a\) where one root is twice the other. Understanding these relationships allows students to solve complex root-related problems more efficiently.

Roots of Equations

The roots of an equation are the solutions that make the equation true. For a quadratic equation, these are the values of \(x\) that satisfy \(ax^2 + bx + c = 0\).

• In a quadratic equation with real coefficients, the roots can be real or complex numbers.
• The nature of the roots depends on the discriminant \(b^2 - 4ac\).

In this exercise, we were given that one root is twice the other. This specific condition guided us in setting up the equations. By letting the roots be \(r\) and \(2r\), it simplified the substitution process when using Vieta’s formulas. Understanding how to derive and work with these conditions is crucial in solving root-based quadratic problems.
Students should appreciate how setting conditions on roots can simplify expression manipulations. It transforms the problem into a more manageable form, which, in this case, leads directly to solving for \(a\).

Product of Roots

The product of roots in a quadratic equation is another key relationship given by Vieta's formulas. Represented by \(r_1 \cdot r_2 = \frac{c}{a}\), where \(c\) and \(a\) are coefficients of the quadratic equation, it provides valuable insights.

• In the given exercise, we represented the product as \(2r^2 = \frac{2}{a^2-5a+3}\).
• This helped in deducing the expression \(r^2 = \frac{1}{a^2-5a+3}\), which was essential for finding the correct value of \(a\).

By understanding the product of the roots, one can identify key characteristics of the parabola associated with the quadratic equation. This relationship aids in analytical assessments without needing graphical or numerical methods.
Grasping this concept not only helps in solving quadratic equations but also in interpreting their geometric and algebraic properties better. Applying these concepts in various scenarios, like this exercise, deepens a student's comprehension of polynomial functions.