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Proof Problem #2

If \(a, b\) are the roots of \(x^4 + x^3 - 1 = 0\) then prove that \(ab\) is a root of \(x^6 + x^4 + x^3 - x^2 - 1 = 0\)

Note by Dev Sharma
2 years, 2 months ago

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Suppose \(a,b,c,d\) are the four roots of \(f(x) = x^{4} + x^{3} - 1.\) Then \(f(x) = (x - a)(x - b)(x - c)(x - d) = \)

\(x^{4} - (a + b + c + d)x^{3} + (ab + ac + ad + bc + bd + cd)x^{2} - (abc + abd + acd + bcd)x + abcd.\)

Comparing like coefficients gives us that \(a + b + c + d = -1, ab + ac + ad + bc + bd + cd = 0,\)

\(abc + abd + acd + bcd = 0\) and \(abcd = -1.\)

Now look at the function \(g(x) = (x - ab)(x - ac)(x - ad)(x - bc)(x - bd)(x - cd).\)

When we expand this expression, we can use the equations involving \(a,b,c,d\) above to establish numerical values for the coefficients of \(g(x).\) For example, the coefficient for \(x^{5}\) will be \(-(ab + ac + ad + bc + bd + cd) = 0,\) and the constant term, i.e., the coefficient of \(x^{0}\), will be \((abcd)^{3} = -1.\) Calculating the other coefficients is quite tedious, (and a bit tricky), but in doing so one will find that \(g(x) = x^{6} + x^{4} + x^{3} - x^{2} - 1,\) and so the product of any two of the roots of the first equation will be a root of the second equation.

Had you already got this far and were having difficulty calculating the coefficients, or is this enough of a hint to help you finish the proof?

Brian Charlesworth - 2 years, 2 months ago

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Nice suggestion...

Dev Sharma - 2 years, 2 months ago

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@Dev Sharma A and b are not the only two roots of the equation.

Mehul Arora - 2 years, 2 months ago

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a and b are two roots. But it has other two root also

Dev Sharma - 2 years, 2 months ago

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Yes, but you should mention that. You should phrase the question like a and b are two roots of the equation........

Mehul Arora - 2 years, 2 months ago

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