Question

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

Answer 1

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

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

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

$2x^2+7x-6=0 \\\ D=7^2-4\cdot2\cdot(-6)=97 \\\ x= \frac{-7\pm \sqrt{97} }{4} \\\ x_1+x_2=\frac{-7+ \sqrt{97} }{4} +\frac{-7- \sqrt{97} }{4} =\frac{-7+ \sqrt{97}-7- \sqrt{97} }{4} =-\frac{14 }{4} =-3.5 \\\ x_1x_2=\frac{-7+ \sqrt{97} }{4} \cdot \frac{-7- \sqrt{97}}{4} =\frac{(-7+ \sqrt{97}) (-7- \sqrt{97})}{4\cdot4}=-\frac{97- 49}{16}=-\frac{48}{16}=-3$

$2x^2+9x+8=0 \\\ D=9^2-4\cdot2\cdot8=17 \\\ x= \frac{-9\pm \sqrt{17} }{4} \\\ x_1+x_2=\frac{-9- \sqrt{17} }{4} +\frac{-9+ \sqrt{17} }{4} =\frac{-9- \sqrt{17}-9+ \sqrt{17} }{4} =-\frac{18 }{4} =- \frac{9}{2} \\\ x_1x_2=\frac{-9- \sqrt{17} }{4} \cdot\frac{-9+ \sqrt{17} }{4} =\frac{(-9- \sqrt{17}) (-9+ \sqrt{17})}{4\cdot4}=-\frac{17- 81}{16}=\frac{64}{16}=4$