Question

найдите сумму квадратов корней уравнения 4x^2-9x-1=0​

Answer 1

$4x^{2} -9x-1=0\\D=(-9)^{2} +4*4*(-1)=97\\x1=\frac{-(-9)+\sqrt{97} }{2*4}=\frac{9}{8}+\frac{1}{8}\sqrt{97}\\x2=\frac{-(-9)-\sqrt{97} }{2*4}=\frac{9}{8}-\frac{1}{8}\sqrt{97}\\(\frac{9}{8}+\frac{1}{8}\sqrt{97})+(\frac{9}{8}-\frac{1}{8}\sqrt{97})=(\frac{9}{8}+\frac{\sqrt{97}}{8})+(\frac{9}{8}-\frac{\sqrt{97}}{8})=\frac{9+\sqrt{97}}{8}+\frac{9-\sqrt{97}}{8}=\frac{9+\sqrt{97}+9-\sqrt{97}}{8}=\frac{18}{8}=\frac{9}{4}=2\frac{1}{4}\\     P.S.PROSHU PROSTIT ZA TO,CHO TAK DOLGO I ZA ETOT UBOGIY TEKST.$