Given that \(\alpha\) is a root of the equation \(x^2-x+3=0\), show that

(a) \(\alpha^3+2\alpha+3=0\)

(b) \(\alpha^4+5\alpha^2+9=0\)

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## Comments

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TopNewesta^3+2a+3=a(a^2+2)+3=a(a^2-a+3+a-3+2)+3=a(0+a-1)+3=a^2-a+3=0 similarly second one can be done

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Since a (using instead of alpha) is a root of x^2 - x + 3 =0 so a satisfies it.hence a^2 - a + 3 =0. Now dividing a^3 + 2a +3 by a^2 - a + 3 we get quotient=a+1& remainder=0. So a^3 +2a+3=(a+1)(a^2 -a+3) + 0 (divident = quotient*divisor + remainder). Since a^2 -a+3=0, so a^3 +2a+3=0.hence the ans.... Similarly we divide a^4 +5a^2 +9 by a^2 -a+3 getting quotient=a^2 +a+3 & remainder =0. Since a^2 -a+3=0 so we get, a^4+ 5a^2 +9 =0...

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Hint:(a) Multiply the first equation by \(x\). Clearly \(\alpha\) is also a root to this equation. Now we substitute \(x^2\) from the original equation.Exercise:Solve (b)Log in to reply

multiply by a. plug in value of a^3.

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You simply need to show that a^2 - a + 3 is a root for both these equations(used a instead of alpha)

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Is there a more efficient way of doing so? (Hint: Manipulate the given equation)

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