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Chapter 3: Problem 223

If \(\alpha, \beta\) are roots of \(x^{2}+p x+q=0\) and \(x^{2 n}+p^{n} x^{n}+q^{n}=0\) and if \((\alpha / \beta)\) is one root of \(\mathrm{x}^{\mathrm{n}}+1+(\mathrm{x}+1)^{\mathrm{n}}=0\) then \(\mathrm{n}\) must be (a) even integer (b) odd integer (c) rational but not integer (d) None of these

Short Answer

Expert verified
(a) even integer

Step by step solution

01

Find the relation between α and β

Since \(\alpha\) and \(\beta\) are roots of the quadratic equation \(x^2+px+q=0\), we have the following relationships from Vieta's formulas: \(\alpha + \beta = -p\) \(\alpha\beta = q\)
02

Analyze the second equation

Given the equation \(x^{2n} + p^n x^n + q^n = 0\), which also has \(\alpha\) and \(\beta\) as roots, we have: \(\alpha^{2n} + p^n\alpha^n + q^n = 0\) \(\beta^{2n} + p^n\beta^n + q^n = 0\) Divide the first equation by \(\alpha^{2n}\) and second equation by \(\beta^{2n}\), we get: \(1 + \frac{p^n\alpha^n}{\alpha^{2n}} + \frac{q^n}{\alpha^{2n}} = 0\) \(1 + \frac{p^n\beta^n}{\beta^{2n}} + \frac{q^n}{\beta^{2n}} = 0\) Subtract the second equation from the first, we have: \(\frac{p^n\alpha^n}{\alpha^{2n}} - \frac{p^n\beta^n}{\beta^{2n}} = \frac{q^n}{\beta^{2n}} - \frac{q^n}{\alpha^{2n}}\)
03

Analyze the third equation and its root

The equation \(x^n + 1 + (x+1)^n = 0\) has a root, say \(r = \frac{\alpha}{\beta}\). We can rewrite the equation as: \((\frac{\alpha}{\beta})^n + 1 + (\frac{\alpha}{\beta} + 1)^n = 0\)
04

Find the relation between the roots

From Step 1, we have \(\alpha + \beta = -p\) and \(\alpha\beta = q\). Since \(r = \frac{\alpha}{\beta}\), we can rewrite the equation from Step 3 as: \((\frac{\alpha}{\beta})^n = -1 - (\frac{\alpha}{\beta} - \frac{\beta}{\alpha})^n\) Now we have a relationship between the roots of the third equation and the first two equations.
05

Determine the property of n

From the equation in Step 4, we can see that if \(n\) is an odd integer, then both terms on the right side will be odd, so their sum will be even, which contradicts the left side, which must always be odd. Therefore, \(n\) cannot be an odd integer. If \(n\) is an even integer, then both terms on the right side will be even, and their sum could potentially be odd, satisfying the left side. Thus, it is possible that \(n\) could be an even integer. So the correct answer is (a) even integer.

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Most popular questions from this chapter

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