Q.Solve the equations

Step 1. Part (i): $6x^4-35x^3+62x^2-35x+6=0$. Coefficients $6,-35,62,-35,6$ are palindromic (Type I). Divide by $x^2$: $6\left(x^2+\dfrac1{x^2}\right)-35\left(x+\dfrac1x\right)+62=0$.

Step 2. Substitute $u=x+\tfrac1x$ (so $x^2+\tfrac1{x^2}=u^2-2$). $6(u^2-2)-35u+62=0 \implies 6u^2-35u+50=0$. $\Delta=1225-1200=25$; $u=\dfrac{35\pm5}{12}$, giving $u=\tfrac{10}3$ or $u=\tfrac52$.

Step 3. Back-substitute each $u$. $u=\tfrac{10}3: x+\tfrac1x=\tfrac{10}3 \implies 3x^2-10x+3=0 \implies (3x-1)(x-3)=0 \implies x=3,\tfrac13$. $u=\tfrac52: x+\tfrac1x=\tfrac52 \implies 2x^2-5x+2=0 \implies (2x-1)(x-2)=0 \implies x=2,\tfrac12$. …

Reciprocal Equations — Type I uses $u=x+1/x$ directly; Type II first peels o …

• Trying $u=x+1/x$ on a Type II equation without first removing the $x=\pm1$ factor …