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Chapter 1: Problem 65

If \(a, b, c, d\) are four consecutive terms of an increasing A.P. then the roots of the equation \((x-a)(x-c)+2(x-b)(x-d)=0\) arc a. non-real complex b. real and equal c. integers d. real and distinct

Short Answer

Expert verified
The roots of the equation are real and distinct, so the answer is (d).

Step by step solution

01

Understanding Arithmetic Progression

In an arithmetic progression (A.P), the difference between consecutive terms is constant. If \(a, b, c, d\) are consecutive terms in such a sequence, then they can be expressed as \(a, a+k, a+2k, a+3k\) respectively, where \(k\) is the common difference.
02

Substitute A.P Terms Into the Equation

Substituting the terms \(a, a+k, a+2k, a+3k\) into the given equation, we have:\[(x-a)(x-(a+2k)) + 2(x-(a+k))(x-(a+3k)) = 0.\] Simplify each pair of terms in the brackets.
03

Simplify the Substituted Equation

Expand and simplify \((x-a)(x-a-2k)\) and \(2(x-a-k)(x-a-3k)\):- \((x-a)(x-a-2k) = x^2 - (2a + 2k)x + a(a+2k)\).- \(2(x-a-k)(x-a-3k) = 2(x^2 - (2a + 3k)x + (a+k)(a+3k))\).Substitute these expansions back into the equation and combine like terms.
04

Combined and Simplify Terms

After combining like terms, the equation is:\[3x^2 - 6kx - 3a^2 - 10ka - 6k^2 = 0.\] Divide each term by 3 to simplify:\[x^2 - 2kx - a^2 - \frac{10ka}{3} - 2k^2 = 0.\]
05

Solve The Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2k\), \(c = -a^2 - \frac{10ka}{3} - 2k^2\). Calculate the discriminant \(b^2 - 4ac\) to determine the nature of the roots.
06

Determine the Nature of Roots

Calculate the discriminant:\(b^2 - 4ac = (2k)^2 - 4(-a^2 - \frac{10ka}{3} - 2k^2) = 4k^2 + 4a^2 + \frac{40ka}{3} + 8k^2\).Since the discriminant is positive, the equation has two distinct real roots.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Consecutive Terms
In an arithmetic progression (A.P.), each term follows a pattern based on a fixed number known as the common difference. Consider four consecutive terms in an A.P., often denoted as \(a, b, c, d\). To represent these in their relationship, you can write them as \(a, a+k, a+2k, a+3k\), where \(k\) is the consistent step from one term to the next.

  • \(a\) - The starting point of the progression.
  • \(a+k\) - The second term, increased by \(k\).
  • \(a+2k\) - The third term, stepping further by \(k\).
  • \(a+3k\) - The fourth term, continuing the sequence.
This understanding of consecutive terms lays the foundation for solving equations derived from such progressions, ensuring you can predict and calculate outcomes efficiently.
Quadratic Equation
Quadratic equations form the heart of many algebraic processes. Standard quadratic form is expressed as \(ax^2 + bx + c = 0\). In the exercise, the equation generated after substituting A.P. terms and simplifying is \(x^2 - 2kx - a^2 - \frac{10ka}{3} - 2k^2 = 0\). A key method to solve it is the quadratic formula:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
This formula delivers the roots' values by considering each coefficient of the quadratic term. Here, \(a = 1\), \(b = -2k\), and \(c = -a^2 - \frac{10ka}{3} - 2k^2\). Understanding the construction of quadratic equations and the process of solving them is vital for analyzing the nature of their solutions.
Discriminant of Quadratic
The discriminant of a quadratic equation, denoted as \(b^2 - 4ac\), holds the secret to the nature of the equation's roots. Let's examine it within the context of our problem:
  • If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
  • If \(b^2 - 4ac = 0\), both roots are real and identical.
  • If \(b^2 - 4ac < 0\), the roots are complex and non-real.
In our exercise, substituting into the discriminant, we find:
  • \(b^2 = 4k^2\)
  • \(-4ac = 4a^2 + \frac{40ka}{3} + 8k^2\)
The calculation reveals a positive value, confirming that the roots are real and distinct. Having a grasp on how the discriminant works is essential for quickly identifying the types of solutions a quadratic equation can yield.

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