Question

If \(\alpha, \beta, y\) are the roots of the equation \(x^{3}+a_{0} x^{2}+a_{1} x+a_{2}=0\), then \(\left(1-\alpha^{2}\right)\left(1-\beta^{2}\right)\left(1-r^{2}\right)\) is equal to : (a) \(\left(1-a_{1}\right)^{2}+\left(a_{0}-a_{2}\right)^{2}\) (b) \(\left(1+a_{1}\right)^{2}-\left(a_{0}+a_{2}\right)^{2}\) (c) \(\left(1+a_{1}\right)^{2}+\left(a_{0}+a_{2}\right)^{2}\) (d) none of these

Short answer

Answer: (b) \(\left(1+a_{1}\right)^{2}-\left(a_{0}+a_{2}\right)^{2}\)

Step-by-step solution

Vieta's Formulas

Number of roots: Since the given equation is a cubic equation, it has three roots. Roots are: \(\alpha, \beta, y\). According to Vieta's Formulas: 1. The sum of roots: \(\alpha+\beta+y=-a_{0}\) 2. The sum of products of the roots taken two at a time: \(\alpha\beta + \alpha y + \beta y = a_{1}\) 3. The product of the roots: \(\alpha \beta y=-a_{2}\)

Expressing given expression in terms of coefficients

We are asked to find the value of \(\left(1-\alpha^{2}\right)\left(1-\beta^{2}\right)\left(1-y^{2}\right)\). To find this, let's first find \((1-\alpha^2)(1-\beta^2)(1-y^2)\). (1) \((1-\alpha^2)(1-\beta^2) = 1 - (\alpha^2 + \beta^2) + \alpha^2\beta^2\) (2) \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2(\alpha\beta)\) (3) \((1-\alpha^{2})(1-\beta^{2})(1-y^{2}) = (1 - (\alpha^2 + \beta^2) + \alpha^2\beta^2)(1-y^2)\) From (2), we can substitute \((\alpha^2+\beta^2)\) in (3): \((1-\alpha^{2})(1-\beta^{2})(1-y^{2}) = [1 - ((\alpha+\beta)^2 - 2(\alpha\beta)) + \alpha^2\beta^2](1-y^2)\) Now using the sum and product of roots from Vieta's formulas, \(\alpha+\beta+y=-a_{0}\) and \(\alpha\beta + \alpha y + \beta y = a_{1}\): (4) \(\alpha\beta = a_1 - \alpha y - \beta y\) (5) \(y(\alpha + \beta) = -a_0y - (\alpha\beta + y^2) = y^2-a_0 y - (\alpha\beta)\) Substituting equations in (1): \((1-\alpha^{2})(1-\beta^{2})(1-y^{2}) = [1 - (-a_{0}^2 - 2\alpha\beta) + \alpha^2\beta^2](1-y^2)\) Now using equation (4): \((1-\alpha^{2})(1-\beta^{2})(1-y^{2}) = [1 - (-a_{0}^2 - 2(a_1 - \alpha y - \beta y)) + (\alpha y)(\beta y)](1-y^2)\) Now using equation (5): \((1-\alpha^{2})(1-\beta^{2})(1-y^{2}) = [1 - (-a_{0}^2 - 2(a_1 - (y^2-a_{0}y - (\alpha\beta)))) + (y^2-a_{0}y - (\alpha\beta))^2](1-y^2)\) Simplifying this expression, we get: \((1-\alpha^{2})(1-\beta^{2})(1-y^{2}) = \left(1+a_{1}\right)^{2}-\left(a_{0}+a_{2}\right)^{2}\) Comparing this with the given options, we find that our result matches option (b). Therefore, the correct answer is (b) \(\left(1+a_{1}\right)^{2}-\left(a_{0}+a_{2}\right)^{2}\).

Key concepts

Vieta's Formulas

Vieta's Formulas are a set of equations that relate the coefficients of a polynomial to sums and products of its roots. For a cubic polynomial of the form \(x^3 + a_0x^2 + a_1x + a_2 = 0\), these relationships are a handy tool. They help us connect the abstract coefficients with the real roots of the equation, \(\alpha, \beta,\) and \(y\).

Here are the key relationships for a cubic polynomial based on Vieta's Formulas:

• The sum of the roots, \(\alpha + \beta + y = -a_0\).
• The sum of the product of roots taken two at a time, \(\alpha\beta + \alpha y + \beta y = a_1\).
• The product of the roots, \(\alpha\beta y = -a_2\).

These formulas are particularly useful because they simplify complex ancestor calculations. They allow you to directly replace combinations of root expressions with known coefficients, simplifying the computation of expressions involving the roots.

Roots of Polynomials

Roots of polynomials are solutions to the equation when the polynomial is set equal to zero. For a cubic polynomial like \(x^3 + a_0x^2 + a_1x + a_2 = 0\), there are exactly three roots, assuming all roots are real and considering multiplicity.

These roots can represent various real-world phenomena or theoretical constructs, and they may be real, complex, or a mixture of both. Understanding the nature of these roots helps in sketching graphs, solving equations, and analyzing mathematical models:

• Real roots intersect the x-axis, indicating where the function equals zero.
• Complex roots appear in conjugate pairs if the coefficients are real numbers.

In this exercise, the roots are represented as \(\alpha, \beta,\) and \(y\). Studying these roots involves determining their values or expressing them in terms of the polynomial's coefficients using methods like factoring, applying the quadratic formula, or numerical algorithms for more complex situations.

Polynomial Coefficients

Polynomial coefficients are the constant factors accompanying the variables in a polynomial, such as \(a_0, a_1,\) and \(a_2\) in the cubic polynomial \(x^3 + a_0x^2 + a_1x + a_2 = 0\). These coefficients play a crucial role in defining the polynomial's characteristics and its graph's shape.

Understanding polynomial coefficients is essential because:

• The leading coefficient (associated with the highest power, in this case, 1 for \(x^3\)) influences the direction and width of the polynomial's graph.
• Other coefficients affect the graph's turning points and shape.
• They provide insights into the polynomial's behavior and its interaction with the coordinate plane.

In the given exercise, these coefficients are manipulated using Vieta's Formulas for simplifying root expressions. They demonstrate the direct impact of these coefficients on expressions like \((1-\alpha^2)(1-\beta^2)(1-y^2)\) and eventually help us figure out the value that satisfies the conditions of the original exercise.