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Let $\alpha_1, \alpha_2, \alpha_3$ be the roots of the polynomial $p(x)=x^3+5x^2+7x+11$. Find a polynomial whose roots are $\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_2+\alpha_3}{2}, \frac{\alpha_1+\alpha_3}{2}.$ Calculating the exact roots doesn't really help because only one root is real. I've tried expanding

$$ \left(x-\frac{\alpha_1+\alpha_2}{2}\right)\left(x-\frac{\alpha_2+\alpha_3}{2}\right)\left(x-\frac{\alpha_1+\alpha_3}{2}\right) $$

but it gets kind of messy. Any "nicer" ideas?

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Let $\alpha,\beta,\gamma$ are the roots of the equation $x^3+5x^2+7x+11=0\;,$ Then $\alpha+\beta +\gamma = -5$

and $\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha = 7$ and $\alpha\cdot \beta \cdot \gamma = -11$

Now we have to calculate a polynomial equation

whose roots are $\displaystyle \left(\frac{\alpha+\beta}{2}\right)\;,\left(\frac{\beta+\gamma}{2}\right)$ and $\displaystyle \left(\frac{\gamma+\alpha}{2}\right).$

So Let $$\displaystyle P = \left(\frac{\alpha+\beta}{2}\right)+\left(\frac{\beta+\gamma}{2}\right)+\left(\frac{\gamma+\alpha}{2}\right)=\alpha+\beta+\gamma = -5$$

and Let $$\displaystyle Q = \left(\frac{\alpha+\beta}{2}\right)\cdot \left(\frac{\beta+\gamma}{2}\right)+\left(\frac{\beta+\gamma}{2}\right)\cdot \left(\frac{\gamma+\alpha}{2}\right)+\left(\frac{\gamma+\alpha}{2}\right)\cdot \left(\frac{\alpha+\beta}{2}\right)$$

So we get $$\displaystyle Q = \frac{1}{4}\left[\alpha^2+\beta^2+\gamma^2+3\alpha\beta+3\beta\gamma+3\gamma\alpha\right] = \frac{1}{4}\left[(\alpha+\beta+\gamma)^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\right]$$

So we get $\displaystyle Q =\frac{1}{4}\left[25+7\right] = 8$

Now Let $$\displaystyle R = \left(\frac{\alpha+\beta}{2}\right)\cdot \left(\frac{\beta+\gamma}{2}\right)\cdot \left(\frac{\gamma+\alpha}{2}\right) = \frac{1}{8}\left[(5+\alpha)\cdot (5+\beta)\cdot (5+\gamma)\right]$$

So we get $$\displaystyle R = -\frac{1}{8}\left[125+5(\alpha\beta+\beta\gamma+\gamma\alpha)+\alpha\beta\gamma+25(\alpha+\beta+\gamma)\right] = -\frac{24}{8} = -3$$

above we used the relation $$\displaystyle \alpha+\beta+\gamma = -5\Rightarrow \alpha+\beta = -5-\alpha$$ etc...

So our polynomial equation is $$\displaystyle x^3-Px^2+Qx+R = 0\;,$$ Now put $P,Q$ and R

So we get a polynomial equation $$\displaystyle x^3+5x^2+8x+3 = 0$$

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  • $\begingroup$ How did you get those first three equations : $\alpha+\beta +\gamma = -5$ , and the next two ? $\endgroup$ – Jean Aug 31 '15 at 1:46
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    $\begingroup$ If $x = \alpha,x=\beta,x=\gamma$ are the roots of $x^3+5x^2+7x+11=0\;,$ Then $(x-\alpha),(x-\beta)$ and $(x-\gamma)$ are three factors of these equation. So $x^3+5x^2+7x+11 = (x-\alpha)(x-\beta)(x-\gamma)$ . Now equating Coefficients. $\endgroup$ – juantheron Aug 31 '15 at 1:50
  • $\begingroup$ I used Vieta's formulas a couple hours after asking this question but I wasn't sure if my solution was correct. Now I see it is! :) @juantheron $\endgroup$ – implicati0n Aug 31 '15 at 12:43
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$\bf{Alternatively::}$ If $x=\alpha\;,x=\beta\;,x=\gamma$ are the roots

of the equation $x^3+5x^2+7x+11=0.\;,$ Then $\alpha+\beta+\gamma = -5$

Now have to find an equation whose roots are $\displaystyle \frac{\alpha+\beta}{2}$ and $\displaystyle \frac{\beta+\gamma}{2}$ and $\displaystyle \frac{\gamma+\alpha}{2}$

Now Using above relation $$\alpha+\beta+\gamma = -5$$

We can write $$\alpha+\beta = -5-\gamma$$ and $$\beta+\gamma = -5-\alpha$$ and $\gamma+\alpha = -5-\beta$

So now we have to find an equation whose roots are $$\displaystyle \frac{-5-\alpha}{2}\;\;,\frac{-5-\beta}{2}\;\;,\frac{-5-\gamma}{2}.$$

so Let $$\displaystyle y = \frac{-5-x}{2}\Rightarrow 2y=-5-x\Rightarrow x=-\left(5+2y\right).$$

Now Put value of $y$ into $x^3+5x^2+7x+11=0\;,$

We get $$\displaystyle -(2y+5)^3+5(2y+5)^2-7(2y+5)+11=0$$

So our polynomial equation is $$(2y+5)^3-5(2y+5)^2+7(2y+5)-11=0$$

So we get $$8y^3+40y^2+64y+24 = 0\Rightarrow y^3+5y^2+8y+3=0$$

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