Problem 19 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If the sum of the roots of the equation \(x^{2}+\) \(p x+q=0\) is three times their difference, (a) \(9 p^{2}=2 q\) (b) \(2 q^{2}=9 p\) (c) \(2 p^{2}=9 q\) (d) \(9 q^{2}=2 p\)

Step-by-Step Solution

Answer

The correct answer is (c): \( 2p^2 = 9q \).

Step 1: Understanding the Relation Between Sum and Difference

Given that the sum of the roots is three times their difference, let the roots be \( \alpha \) and \( \beta \). According to Vieta’s formulas, the sum of the roots \( \alpha + \beta = -p \) and the product \( \alpha \beta = q \). The problem states \( \alpha + \beta = 3(\alpha - \beta) \).

Step 2: Expressing the Angle of Roots' Difference

Rewrite the equation \( \alpha + \beta = 3(\alpha - \beta) \) to express \( \beta \) in terms of \( \alpha \). Expand and solve: \( \alpha + \beta = 3\alpha - 3\beta \rightarrow \alpha + \beta + 3\beta = 3\alpha \rightarrow \alpha + 4\beta = 3\alpha \).

Step 3: Refine and Solve the Equation

From \( \alpha + 4\beta = 3\alpha \), we simplify to obtain \( 4\beta = 2\alpha \). Dividing throughout by 2 gives \( 2\beta = \alpha \). Thus, \( \alpha = 2\beta \).

Step 4: Substitute Back Into Sum and Product Relations

Substitute \( \alpha = 2\beta \) into the sum relation: \( 2\beta + \beta = -p \Rightarrow 3\beta = -p \). Substitute into the product relation: \( 2\beta \cdot \beta = q \Rightarrow 2\beta^2 = q \).

Step 5: Examine Given Options Against Relation

Now express \( p \) and \( q \) in terms of \( \beta \): from \( 3\beta = -p \), we obtain \( \beta = \frac{-p}{3} \). Substitute this into \( 2\beta^2 = q \) gives \( 2(\frac{-p}{3})^2 = q \), which simplifies to \( 2\frac{p^2}{9} = q \). Then \( q = \frac{2p^2}{9} \) which checks against option (c): \( 2p^2 = 9q \).

Step 6: Conclusion

Since the derived equation \( 2p^2 = 9q \) matches option (c), the correct answer is verified.

Key Concepts

Roots of Quadratic Equation

Quadratic equations are fundamental expressions of the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants. The solutions to these equations are known as the roots. These roots represent the values of \( x \) for which the quadratic expression equals zero.

To find the roots of a quadratic equation, one can use several methods: the quadratic formula, factoring, or completing the square. The quadratic formula is given by:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula provides the solutions directly based on the coefficients \( a \), \( b \), and \( c \). The discriminant, \( b^2 - 4ac \), determines the nature of the roots, indicating whether they are real or complex.

Vieta's Formulas

Vieta's formulas offer valuable insights into the relationships between the coefficients of a polynomial and its roots. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), if \( \alpha \) and \( \beta \) are the roots, then Vieta's formulas state:

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

These relationships permit us to express the roots in terms of the coefficients, providing a powerful tool for solving quadratic equations. Essentially, they link the behavior of the roots directly to the structure of the equation. This can significantly simplify the solving process by transforming a potentially complex root-finding problem into a matter of substitution and algebraic manipulation.

Sum and Product of Roots

Understanding the sum and product of the roots allows for a deeper grasp of quadratic equations. From Vieta's Theorems, in the equation \( ax^2 + bx + c = 0 \), the sum \( \alpha + \beta \) and the product \( \alpha \beta \) are directly related to the equation's coefficients:

• Sum: \( \alpha + \beta = -\frac{b}{a} \)
• Product: \( \alpha \beta = \frac{c}{a} \)

These expressions offer not only a quick way to verify solutions but also to reconstruct the original equation from its roots. In practical terms, knowing just the sum and product of roots can tell us a great deal about the equation without requiring actual root values. This has wide applications in algebra and even in problem-solving scenarios in competitions.

Solving Quadratic Equations

Solving quadratic equations is the process of finding the roots that satisfy the equation \( ax^2 + bx + c = 0 \). Depending on the context, you may use different methods: factoring, using the quadratic formula, or completing the square.

• **Factoring:** Suitable for simple quadratics where the factors are apparent.
• **Quadratic Formula:** \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which works universally.
• **Completing the Square:** Transforms the equation into a perfect square trinomial.

Each of these methods highlights different facets of quadratic equations, offering a range of tools to find solutions effectively. A special technique involves first leveraging Vieta's formulas to relate the sums and products of roots directly to broaden the approach, as seen in many algebraic problems.