Question

Решите квадрат уравнение (20-22): 5.20. a) x2-x- 90 = 0; B) 4x² - x - 3 = 0​​

Answer 1

Ответ:

a) $x=-9,\ x=10$

б) $x=- \frac{ 3 }{ 4 },\ x=1$

Объяснение:

a)

$x^{2} - x - 90 =0\\\\ a=1 ,\ \ b=-1 ,\ \ c=-90\\\\D = b^2 - 4ac = ( - 1)^2 - 4\cdot1\cdot( - 90) = 1 + 360 = 361\\\\\\\sqrt{D} =\sqrt{361} = 19\\\\x_1=\frac{-b-\sqrt{D}}{2a}=\frac{1-19}{2\cdot1}=\frac{-18 }{2 }=-9\\\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{1+19}{2\cdot1}=\frac{20}{2}=10$

б)

$4x^{2} - x - 3 =0\\\\ a=4 ,\ \ b=-1 ,\ \ c=-3\\\\D = b^2 - 4ac = ( - 1)^2 - 4\cdot4\cdot( - 3) = 1 + 48 = 49\\\\\\\sqrt{D} =\sqrt{49} = 7\\\\x_1=\frac{-b-\sqrt{D}}{2a}=\frac{1-7}{2\cdot4}=\frac{-6 }{8 }=- \frac{ 3 }{ 4 } \\\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{1+7}{2\cdot4}=\frac{8}{8}=1$

Answer 2

$\displaystyle\bf\\1)\\\\x^{2} -x-90=0\\\\D=(-1)^{2} -4\cdot(-90)=1+360=361=19^{2} \\\\\\x_{1} =\frac{1-19}{2} =-\frac{18}{2} =-9\\\\\\x_{2} =\frac{1+19}{2} =\frac{20}{2} =10\\\\\\Otvet \ : \ -9 \ ; \ 10\\\\2)\\\\4x^{2} -x-3=0\\\\D=(-1)^{2} -4\cdot 4\cdot(-3)=1+48=49=7^{2} \\\\\\x_{1} =\frac{1-7}{8} =-\frac{6}{8} =-\frac{3}{4} =-0,75\\\\\\x_{2} =\frac{1+7}{8} =\frac{8}{8} =1\\\\\\Otvet \ : \ -0,75 \ ; \ 1$