Given that α and β are the roots of x2−2x−1=0, find 5α4+12β3.
Solution — click to reveal
If x satisfies x2−2x−1=0, then x2x3x4=2x+1,=x(2x+1)=2x2+x=2(2x+1)+x=5x+2,=x(5x+2)=5x2+2x=5(2x+1)+2x=12x+5.Hence, 5α4+12β3=5(12α+5)+12(5β+2)=60α+25+60β+24=60(α+β)+49=60⋅2+49=169.