Problem 142 - A Complete Resource Book in Mathematics for JEE Main 2017

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: Identify Relationship Between Roots

Let the roots of the quadratic equation be \( r \) and \( 2r \), where one root is twice the other. We know from the relation of roots that \( r + 2r = -\frac{b}{a} \) and \( r \times 2r = \frac{c}{a} \) for an equation of the form \( ax^2 + bx + c = 0 \).

Step 2: Write Equations Using Relationship

For the given equation \((a^2-5a+3)x^2 + (3a-1)x + 2 = 0\), the sum and product of roots are:Sum of roots: \( r + 2r = 3r = -\frac{3a-1}{a^2-5a+3} \)Product of roots: \( r \times 2r = 2r^2 = \frac{2}{a^2-5a+3} \)

Step 3: Express r in Terms of a

From the sum of roots, express \( r \):\[ r = \frac{-(3a-1)}{3(a^2-5a+3)} \]

Step 4: Substitute Expression into Product Equation

Substitute \( r = \frac{-(3a-1)}{3(a^2-5a+3)} \) into the product equation:\[2 \left(\frac{-(3a-1)}{3(a^2-5a+3)}\right)^2 = \frac{2}{a^2-5a+3}\]

Step 5: Equate Equations and Simplify

Equating gives:\[2 \frac{(3a-1)^2}{9(a^2-5a+3)^2} = \frac{2}{a^2-5a+3}\]Solving this, you multiply both sides by \( 9(a^2-5a+3) \) to eliminate fractions, simplifying to:\[(3a-1)^2 = 9(a^2-5a+3)\]Expand it to:\[9a^2 - 6a + 1 = 9a^2 - 45a + 27\]

Step 6: Solve Simplified Equation

Cancel \(9a^2\) from both sides:\[-6a + 1 = -45a + 27\]Rearrange to find \(a\):\[39a = 26\]\[a = \frac{26}{39} = \frac{2}{3}\]

Step 7: Verify Result

Substituting \(a = \frac{2}{3}\) back into the conditions of the problem confirms that this value satisfies both the sum and product relationships. Therefore, the correct answer is consistent with this calculated result.

Key Concepts

Roots of Quadratic Equation

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). This standard form helps us explore the concepts of roots, which are the solutions \((x)\) that satisfy the equation. The roots are essential because they illustrate where the graph of the quadratic equation intersects the x-axis.
In this specific problem, the roots are related by a specific condition: one root is twice the other, or if we call the roots \(r\) and \(2r\). This relationship adds a layer of complexity that requires using algebra to express these roots in terms of the coefficients \(a, b,\) and \(c\).
When handling quadratic equations, you can use various methods to find the roots, such as factoring, completing the square, or using the quadratic formula. In some problems, like this one, you may also use the relationship between the roots to find specific solutions or relationships that satisfy the given conditions.

Sum and Product of Roots

The fundamental properties of any quadratic equation involve the sum and product of its roots. For a quadratic equation \(ax^2 + bx + c = 0\), the sum of its roots is given by \(-\frac{b}{a} \), and their product is \(\frac{c}{a}\). These relationships are incredibly helpful for creating equations when roots have particular conditions.
In our problem, where the roots are \(r\) and \(2r\), we have specific equations:

• Sum of the roots: \(r + 2r = 3r = -\frac{b}{a}\).
• Product of the roots: \(r \times 2r = 2r^2 = \frac{c}{a}\).

These relationships allow us to express one root in terms of the coefficient \(a\) of the given quadratic equation and then use this to form further equations that help us solve for the desired values of \(a\).
Using these properties in certain algebraic manipulations can greatly help in reducing the complexity of the problem.

Equation Relationships

The relationship between the roots and the coefficients in a quadratic equation is crucial to solving problems like ours, where conditions are placed on the roots themselves. By expressing one root in terms of the equation's coefficients, you can set up another equation to solve for unknown parameters.
When solving the given problem, we expressed the roots \(r\) and \(2r\) based on the known relationships of their sum and product to the coefficients of the quadratic equation:

• Substitute these expressions into the product condition.
• Simplify to consolidate all variables.
• Verify the results satisfy both conditions of sum and product.

This translates algebraic manipulation into finding specific values such as \(a\), which satisfies both conditions outlined in the problem.
This process may seem complex at first, but with practice, understanding the interconnectedness of these algebraic expressions makes solving quadratic equation problems much easier.