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Let p(x) be a quadratic polynomial with integer coefficients which has \(4-\sqrt{11}\) as a root. Compute \(\frac{p(3)}{p(4)}\).

 May 28, 2019

Hi Lightning  laugh


\(P(x)=ax^2+bx+c\\ P(4-\sqrt{11})=a(4-\sqrt{11})^2+b(4-\sqrt{11})+c=0\\\qquad \text{Where a,b and c are integers}\\ a(4-\sqrt{11})^2+b(4-\sqrt{11})+c=0\\ a(16+11-8\sqrt{11})+b(4-\sqrt{11})+c=0\\ a(27-8\sqrt{11})+b(4-\sqrt{11})+c=0\\ 27a+4b=-c\quad(1)\\ -8a\sqrt{11}-b\sqrt{11}=0\\ -8a-b=0\\ b=-8a \quad (2)\\ \text{sub two into one}\\ 27a+4*-8a=-c\\ -5a=-c\\ c=5a\)


So the equation becomes

\(P(x)=ax^2-8ax+5a\\ P(x)=a(x^2-8x+5)\\ \)


Find P(3) and P(4) and do the division.


You can finish it   wink


(oh, you do need to check my working, I have not done so. )

 May 28, 2019

I'd make this a bit simpler.


\(\text{We can always choose $P(x)$ to be a monic polynomial}\\ \text{Integer coefficients means $4-\sqrt{11}$ is a root $\Rightarrow 4+\sqrt{11}$ is a root}\\ \text{Thus we have}\\ P(x) = \left(x- 4+\sqrt{11}\right)\left(x-4-\sqrt{11}\right)=x^2-8x+5\)

Rom  May 28, 2019
edited by Rom  May 28, 2019

That is much nicer Rom :)

Melody  May 28, 2019

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