Question

If the roots of \(x^{3}-3 x^{2}+p x+1=0\), are in arithmetic progression, then the sum of squares of the langest and the smallest roots is (a) 3 (b) 5 (c) 6 (d) 10 .

Short answer

The sum of the squares of the largest and smallest roots is 6.

Step-by-step solution

Understand Arithmetic Progression of Roots

If the roots of the cubic equation \(x^3 - 3x^2 + px + 1 = 0\) are in arithmetic progression, denote the roots as \(a-d\), \(a\), and \(a+d\). Here, the middle root \(a\) ensures the other two roots differ symmetrically by \(d\).

Apply Sum of Roots (Vieta's Formula)

For a cubic polynomial \(ax^3 + bx^2 + cx + d = 0\), the sum of the roots \(r_1 + r_2 + r_3 = -\frac{b}{a}\). Given the current polynomial, \(r_1 + r_2 + r_3 = a-d + a + a+d = 3a = 3\). This simplifies to \(a = 1\).

Calculate Roots with the Middle Value

Since \(a = 1\), the three roots are \(1-d\), \(1\), and \(1+d\).

Use Product of Roots (Vieta's Formula) for Parameter p

The product of the roots \((1-d) \cdot 1 \cdot (1+d) = 1-d^2 = -1\), which gives the equation \(d^2 = 2\).

Substitute Roots and Calculate Required Sum of Squares

Calculating the sum of squares of the smallest and largest roots: \((1-d)^2 + (1+d)^2 = (1)^2 - 2\cdot 1 \cdot d + d^2 + (1)^2 + 2 \cdot 1 \cdot d + d^2\) simplifies to \(2 + 2d^2\). Substitute \(d^2 = 2\) from Step 4: \(2 + 2 \times 2 = 6\).

Check the Options and Conclude

The correct answer matches option (c), which is 6.

Key concepts

Arithmetic Progression

Arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the common difference, often represented by "d." For example, if the roots of a cubic equation are in AP, we can represent them as \( a-d, a, \) and \( a+d \). These roots are set apart by the common difference \( d \) and centered around the middle root \( a \). This concept is particularly helpful in simplifying problems, as it allows us to express relationships between numbers using algebraic expressions.

In the context of our exercise, knowing that the roots are in AP helps us apply Vieta's formulas more easily by linking the roots mathematically. Understanding AP can provide a structured way to approach complex mathematical problems and helps visualize relationships between terms.

Cubic Equations

A cubic equation is a polynomial equation of degree three, which generally takes the form \( ax^3 + bx^2 + cx + d = 0 \). Each cubic equation can have up to three roots, which can be all real or a combination of real and complex numbers. Solving cubic equations typically involves finding these roots, which are the values of \( x \) that make the equation true.

In the exercise, we focus on a specific cubic equation \( x^3 - 3x^2 + px + 1 = 0 \), where we are tasked with identifying roots that satisfy certain conditions. By manipulating the equation and applying algebraic techniques like factoring or using Vieta’s formulas, we can derive useful information about the roots. Cubic equations play a crucial role across various fields of science and engineering as they model various natural phenomena.

Vieta's Formulas

Vieta's formulas provide a powerful tool for working with polynomial equations, linking coefficients to sums and products of the roots. For a cubic equation \( ax^3 + bx^2 + cx + d = 0 \), Vieta's formulas imply:

• Sum of the roots, \( r_1 + r_2 + r_3 = -\frac{b}{a} \)
• Sum of the product of roots taken two at a time, \( r_1r_2 + r_2r_3 + r_3r_1 = \frac{c}{a} \)
• Product of the roots, \( r_1r_2r_3 = -\frac{d}{a} \)

In our problem, since the roots \( a-d, a, \) and \( a+d \) are in arithmetic progression, Vieta’s formulas allow us to calculate the sum of the roots and connect it to the quadratic coefficient. These formulas simplify the process of solving cubic polynomials by transforming them into solvable algebraic equations.

Sum of Squares

The "sum of squares" concept is used to calculate the sum of the squared values of specific numbers. In mathematical terms, for any numbers \( x \) and \( y \), the sum of squares is \( x^2 + y^2 \). This is a common operation in statistics and various fields of mathematics to measure variations and analyze data.

In our exercise, we apply this concept to the roots of the polynomial equation to find the sum of squares of the smallest and largest roots. Knowing that the roots are \( 1-d \) and \( 1+d \), we compute \((1-d)^2 + (1+d)^2\). Simplifying this expression involves expanding the squares and substituting known values like \( d^2 = 2 \). Understanding how to work with sums of squares is valuable for algebra and calculus, as it frequently arises in optimization problems and error measurements.