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Exercise 4(d) · Q6

Q.Solve the reciprocal equation x4−2x3−x2−2x+1=0x^4-2x^3-x^2-2x+1=0.

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Step 1. The coefficients 1,−2,−1,−2,11,-2,-1,-2,1 read the same forwards and backwards, so this is a class-one reciprocal equation of even degree 44.

Step 2. Since x=0x=0 is clearly not a root, divide throughout by x2x^2:

x2−2x−1−2x+1x2=0.x^2-2x-1-\frac2x+\frac1{x^2}=0.

Step 3. Group the symmetric pairs: (x2+1x2)−2(x+1x)−1=0\Big(x^2+\dfrac1{x^2}\Big)-2\Big(x+\dfrac1x\Big)-1=0.

Step 4. Let t=x+1xt=x+\dfrac1x, so x2+1x2=t2−2x^2+\dfrac1{x^2}=t^2-2. The equation becomes

(t2−2)−2t−1=0 ⟹ t2−2t−3=0 ⟹ (t−3)(t+1)=0,(t^2-2)-2t-1=0\ \Longrightarrow\ t^2-2t-3=0\ \Longrightarrow\ (t-3)(t+1)=0,

so t=3t=3 or t=−1t=-1. …

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