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Chapter 3: Problem 240

If \(\alpha \& \beta\) are roots of equation \(\mathrm{x}^{2}+\mathrm{x}+1=0\) then the equation whose roots are \(\alpha^{19} \& \alpha^{7}\) is \(\ldots \ldots \ldots\) (a) \(x^{2}-x+1=0\) (b) \(x^{2}+x+1=0\) (c) \(x^{2}+x+3=0\) (d) \(x^{2}-x+3=0\)

Short Answer

Expert verified
The equation whose roots are \(\alpha^{19}\) and \(\alpha^{7}\) is \(x^2 -(\alpha^{19}+\alpha^7)x+\alpha^{26}=0\), which is equivalent to option (b): \(x^2 + x + 1=0\).

Step by step solution

01

Sum and product of roots for given equation

\(S_\alpha = \alpha + \beta = -\frac{1}{1} = -1\) and \(P_\alpha = \alpha \beta = \frac {1}{1} = 1\) #Step 2: Find expressions for the desired roots# Using the equation whose roots are \(\alpha^{19}\) and \(\alpha^{7}\), we can rewrite using the given original roots: Sum of new roots: \(S_{new} = \alpha^{19} + \alpha^{7}\) Product of new roots: \(P_{new} = \alpha^{19} \cdot \alpha^{7} = \alpha^{26}\) Note: Since \(\alpha\beta = \frac{a_2}{a_0}=1\) (as computed in step 1), we can let \(\alpha=\frac{1}{\beta}\). #Step 3: Transform expressions for new roots using the properties of \(\alpha\) and \(\beta\)#
02

New sum and product using properties of the roots

Sum of new roots: \(S_{new} = \left(\frac{1}{\beta}\right)^{19} + \left(\frac{1}{\beta}\right)^{7}\) Product of new roots: \(P_{new} = \left(\frac{1}{\beta}\right)^{26}\) #Step 4: Use Vieta's Formula to find the equation with transformed roots# Recall that if \(x^2 + bx + c=0\) has roots \(r\) and \(s\), then \(b=-(r+s)\) and \(c=rs\) Therefore, we obtain the new equation as follows:
03

Find the new equation using Vieta's Formula

\(x^2 -\left(\left(\frac{1}{\beta}\right)^{19} + \left(\frac{1}{\beta}\right)^{7}\right)x + \left(\frac{1}{\beta}\right)^{26}=0\) #Step 5: Substitute back the value for \(\beta\)# Recall that \(\beta=\frac{1}{\alpha}\). So, substituting in the expression above, we get: \(x^2 -\left(\alpha^{19} + \alpha^{7}\right)x + \alpha^{26}=0\) This is the equation we are looking for, and it is equivalent to option (b): \(x^2 + x + 1=0\).

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