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

**\(x^4 + x^3 - 1 = 0\)** then prove that \(ab\) is a root of **\(x^6 + x^4 + x^3 - x^2 - 1 = 0\)**

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestSuppose \(a,b,c,d\) 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 · 1 year, 5 months ago

Log in to reply

– Dev Sharma · 1 year, 5 months ago

Nice suggestion...Log in to reply

@Dev Sharma A and b are not the only two roots of the equation. – Mehul Arora · 1 year, 6 months ago

Log in to reply

– Dev Sharma · 1 year, 5 months ago

a and b are two roots. But it has other two root alsoLog in to reply

– Mehul Arora · 1 year, 5 months ago

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