Question

Дискриминант

$0,15x^2-3x+7,65=0$
$-\frac{6}{13}x^2 -6x-18 \frac{6}{13} =0$

Answer 1

Ответ:

$1)\ \ 0,15x^2-3x+7,65=0\ \ \ \to \ \ \ a=0,15\ \ ,\ b=-3\ ,\ c=7,65\\\\D=b^2-4ac=9-4\cdot 0,15\cdot 7,65=9-4,59=4,41=2,1^2\\\\x=\dfrac{-b\pm \sqrt{D}}{2a}\ \ ,\ \ x_1=\dfrac{3-2,1}{0,3}=\dfrac{0,9}{0,3}=3\ ,\ \ x_2=\dfrac{3+2,1}{0,3}=\dfrac{5,1}{0,3}=17\\\\Otvet:\ x_1=3\ ,\ x_2=17\ .$

$2)\ \ -\dfrac{6}{13}x^2-6x-18\dfrac{6}{13}=0\ \Big|\cdot (-13)\ \ \to \ \ 6x^2+78x+240=0\ \Big|:2\ ,\\\\3x^2+39x+120=0\ \ \to \ \ a=3\ \ ,\ b=39\ ,\ c=120\\\\D=b^2-4ac=39^2-4\cdot 3\cdot 120=1521-1440=81\\\\x_1=\dfrac{-39-9}{6}=-8\ ,\ \ x_2=\dfrac{-39+9}{6}=-5\\\\Otvet:\ x_1=-8\ ,\ x_2=-5\ .$