Question

помогите решить пожалуйста

Answer 1

$x^{2} -2x-3=0\\x_{1} +x_{2}=2\\x_{1} *x_{2}=-3\\ x_{1} =3\\ x_{2}=-1$

Ответ: -1; 3

$3x^{2} -14x+16=0\\D=b^{2}-4ac\\ D=196-4*3*16=4\\x_{1}=\frac{-b+\sqrt{D} }{2a} \\x_{1}=\frac{14+2}{2*3}=2\frac{4}{6}=2\frac{2}{3} \\x_{2}=\frac{-b-\sqrt{D} }{2a} \\ x_{2}=\frac{14-2}{6}=2$

Ответ: 2; $2\frac{2}{3}$

$7z^{2}-20z+14=0\\D=b^{2}-4ac\\D=400-4*7*14=8\\z_{1}=\frac{-b+\sqrt{D} }{2a} \\z_{1}=\frac{20+\sqrt{8} }{2*7}=\frac{20+\sqrt{2*4} }{14} =\frac{20+2\sqrt{2} }{14}=\frac{2(10+\sqrt{2)} }{14}=\frac{10+\sqrt{2} }{7}\\z_{2}=\frac{-b-\sqrt{D} }{2a} \\ z_{2}=\frac{20-2\sqrt{2} }{14}=\frac{2(10-\sqrt{2}) }{14}=\frac{10-\sqrt{2} }{7}$

Ответ: $\frac{10-\sqrt{2} }{7} ; \frac{10+\sqrt{2} }{7}$

$p^{2}-10p-25=0\\D=b^{2}-4ac\\ D=100-4*(-25)=200\\p_{1}=\frac{-b+\sqrt{D} }{2a} \\p_{1}=\frac{10+\sqrt{200} }{2}=\frac{10+\sqrt{100*2} }{2}=\frac{10+10\sqrt{2} }{2}=\frac{2(5+5\sqrt{2})}{2}=5+5\sqrt{2}\\ p_{2}=\frac{-b-\sqrt{D} }{2a}\\ p_{2}=\frac{10-10\sqrt{2} }{2}=\frac{2(5-5\sqrt{2})}{2}=5-5\sqrt{2}$

Ответ:$5-5\sqrt{2}; 5+5\sqrt{2}$

$y^{2}-10y+25=0\\ y_{1}+y_{2}=10\\y_{1}*y_{2}=25\\y=5$

Ответ:5