If α, β and γ are the zeros of the polynomial f(x) = ax 3+ bx 2+ cx + d, then α 2+ β 2+ γ 2is equal to
Understanding the Problem
The question asks us to find the value of the sum of squares of the zeros of a cubic polynomial. We are given the polynomial \(f(x) = ax^3 + bx^2 + cx + d\), and its zeros are denoted by \(\alpha\), \(\beta\), and \(\gamma\). We need to find the value of \(\alpha^2 + \beta^2 + \gamma^2\) in terms of the coefficients a, b, c, and d.
To solve this, we will use the relationships between the zeros and the coefficients of a polynomial, commonly known as Vieta's formulas.
Applying Vieta's Formulas for Cubic Polynomials
For a cubic polynomial \(ax^3 + bx^2 + cx + d = 0\), Vieta's formulas provide the following relationships between the zeros (\(\alpha, \beta, \gamma\)) and the coefficients (a, b, c, d):
- Sum of the zeros: \(\alpha + \beta + \gamma = -\frac{b}{a}\)
- Sum of the product of the zeros taken two at a time: \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\)
- Product of the zeros: \(\alpha\beta\gamma = -\frac{d}{a}\)
Calculating the Sum of Squares of Zeros
We are interested in finding the value of \(\alpha^2 + \beta^2 + \gamma^2\). We know a standard algebraic identity that relates the sum of squares to the square of the sum and the sum of pairwise products:
\[(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]
We can rearrange this identity to express the sum of squares:
\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]
Now, we can substitute the expressions from Vieta's formulas into this equation:
Substitute \(\alpha + \beta + \gamma = -\frac{b}{a}\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\):
\[\alpha^2 + \beta^2 + \gamma^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right)\]
Simplify the expression:
\[\alpha^2 + \beta^2 + \gamma^2 = \frac{b^2}{a^2} - \frac{2c}{a}\]
To combine these terms, we find a common denominator, which is \(a^2\):
\[\alpha^2 + \beta^2 + \gamma^2 = \frac{b^2}{a^2} - \frac{2c \cdot a}{a \cdot a}\]
\[\alpha^2 + \beta^2 + \gamma^2 = \frac{b^2 - 2ac}{a^2}\]
Final Result
The value of \(\alpha^2 + \beta^2 + \gamma^2\) is \(\frac{b^2 - 2ac}{a^2}\).
Revision Table: Key Formulas for Cubic Polynomials
| Polynomial | Zeros | Sum of Zeros | Sum of Pairwise Products of Zeros | Product of Zeros |
|---|---|---|---|---|
| \(ax^3 + bx^2 + cx + d\) | \(\alpha, \beta, \gamma\) | \(\alpha + \beta + \gamma = -\frac{b}{a}\) | \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\) | \(\alpha\beta\gamma = -\frac{d}{a}\) |
Additional Information: Generalization of Vieta's Formulas
Vieta's formulas can be applied to polynomials of any degree. For a polynomial of degree \(n\):
\[P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\]
with zeros \(r_1, r_2, \dots, r_n\), the sum of the products of the zeros taken \(k\) at a time is given by:
\[\sum_{1 \le i_1 < i_2 < \dots < i_k \le n} r_{i_1} r_{i_2} \dots r_{i_k} = (-1)^k \frac{a_{n-k}}{a_n}\]
For a cubic polynomial (\(n=3\)), this gives:
- \(k=1\): Sum of zeros = \((-1)^1 \frac{a_{3-1}}{a_3} = -\frac{a_2}{a_3}\) (Matches \(-b/a\))
- \(k=2\): Sum of pairwise products = \((-1)^2 \frac{a_{3-2}}{a_3} = +\frac{a_1}{a_3}\) (Matches \(c/a\))
- \(k=3\): Product of zeros = \((-1)^3 \frac{a_{3-3}}{a_3} = -\frac{a_0}{a_3}\) (Matches \(-d/a\))
This generalized form shows the pattern for any degree polynomial and its zeros.
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