Question

Решите уравнения(8 класс)

Answer 1

1.

$\displaystyle (3x-1)^2+15x=11\\9x^2-6x+1+15x=11\\9x^2+9x-10=0\\D=b^2-4ac=9^2-4\cdot9\cdot (-10)=441\\x_{1}=\frac{-b+\sqrt{D}}{2a} =\frac{-9+21}{18}=\frac{12}{18}=\frac{2}{3} \\x_{2}=\frac{-b+\sqrt{D}}{2a}=\frac{-9-21}{18} =-\frac{30}{18} =-\frac{5}{3}$

Ответ: $x_{1}=\frac{2}{3}; \;x_{2} =-\frac{5}{3} .$

2.

$\displaystyle x^2-2\sqrt{3}x+2=0\\D=b^2-4ac=(-2\sqrt{3})^2-4\cdot1\cdot2=12-8=4\\x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}=\frac{2\sqrt{3}\pm2 }{2} =\frac{2(\sqrt{3}\pm1)}{2}=\sqrt{3}\pm1$

Ответ: $x_{1,2}=\sqrt{3}\pm1.$

Answer 2

$4)\; \; (3x-1)^2+15x=11\\\\9x^2-6x+1+15x-11=0\\\\9x^2+9x-10=0\\\\D=9^2+4\cdot 9\cdot 10=441\; \; ,\; \; \sqrt{D}=\sqrt{441}=21\\\\x_1=\frac{-9-21}{18}=-\frac{30}{18}=-\frac{5}{3}\\\\x_2=\frac{-9+21}{18}=\frac{12}{18}=\frac{2}{3}$

$5)\; \; x^2-2\sqrt3x+2=0\\\\D/4=(\frac{b}{2})^2-ac=(\sqrt3)^2-2=3-2=1\\\\x_1=\frac{-\frac{b}{2}-\sqrt{D}}{a}=\frac{\sqrt3-1}{1}=\sqrt3-1\\\\x_2=\frac{-\frac{b}{2}+\sqrt{D}}{a}=\frac{\sqrt3+1}{1}=\sqrt3+1$