Question

Номер 581.......................………………………

Answer 1

$ax^2+bx+c=0 \ D=b^2-4ac \ x= frac{-bpm sqrt{D} }{2a} \\ ax^2+2kx+c=0 \ D_1=k^2-ac \ x= frac{-kpm sqrt{D_1} }{a}$

$x^2-2x-9=0 \ D_1=1^2-(-9)=10 \ x= 1pm sqrt{10} \ x_1+x_2=1+ sqrt{10} +1- sqrt{10} =2=-(-2) \ x_1x_2=(1+ sqrt{10} )(1- sqrt{10})=1-10=-9$

$3x^2-4x-4=0 \ D_1=2^2-3cdot(-4)=16 \ x_1= frac{2+4 }{3}=2 \ x_2= frac{2-4 }{3}=- frac{2}{3} \ x_1+x_2=2-frac{2}{3} =1frac{1}{3} =frac{4}{3} =-(-frac{4}{3} ) \ x_1x_2=2cdot(-frac{2}{3} )=-frac{4}{3}$

$2x^2+7x-6=0 \ D=7^2-4cdot2cdot(-6)=97 \ x= frac{-7pm sqrt{97} }{4} \ x_1+x_2=frac{-7+ sqrt{97} }{4} +frac{-7- sqrt{97} }{4} =frac{-7+ sqrt{97}-7- sqrt{97} }{4} =-frac{14 }{4} =-3.5 \ x_1x_2=frac{-7+ sqrt{97} }{4} cdot frac{-7- sqrt{97}}{4} =frac{(-7+ sqrt{97}) (-7- sqrt{97})}{4cdot4}=-frac{97- 49}{16}=-frac{48}{16}=-3$

$2x^2+9x+8=0 \ D=9^2-4cdot2cdot8=17 \ x= frac{-9pm sqrt{17} }{4} \ x_1+x_2=frac{-9- sqrt{17} }{4} +frac{-9+ sqrt{17} }{4} =frac{-9- sqrt{17}-9+ sqrt{17} }{4} =-frac{18 }{4} =- frac{9}{2} \ x_1x_2=frac{-9- sqrt{17} }{4} cdotfrac{-9+ sqrt{17} }{4} =frac{(-9- sqrt{17}) (-9+ sqrt{17})}{4cdot4}=-frac{17- 81}{16}=frac{64}{16}=4$