Question

Решите систему уравнений { х^2-у= -1 , х+у=1

Answer 1

$\left \{ {{ x^{2} -y=-1} \atop {x+y=1}} \right. \left \{ {{x^{2} -y=-1} \atop {y=1-x}} \right.$

$x^{2} -(1-x)=-1$
$x^{2} -1+x=-1$
$x^{2} +x=0$
$a=1; b=1;c=0$
$D= b^{2} -4ac$
$D=1^{2} -4*1*0=1-0=1$
$\sqrt{D} =1$
$x= \frac{-b+-\sqrt{D}}{2a}$
$x_{1} = \frac{-1+1}{2*1} = \frac{0}{2} =0$
$x_{2} = \frac{-1-1}{2*1}= \frac{-2}{2} =-1$

$y_{1} =1-0=1$
$y_{2} =1-(-1)=1+1=2$

Ответ: $(0;1);(-1;2)$