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# Problem 4: Symmetrical Properties of Roots

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

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

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

Note by Victor Loh
3 years, 4 months ago

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a^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

- 3 years ago

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...

- 3 years, 4 months ago

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)

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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

- 3 years, 4 months ago

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

- 3 years, 4 months ago