Question

Решите систему уравнений: {у=2х+1 {y^2-10х-11=0

Answer 1

$\left \{ {{y=2x+1} \atop {y^2-10x-11=0}} \right. \left \{ {{y=2x+1} \atop {(2x+1)^2-10x-11=0}} \right. \left \{ {{y=2x+1} \atop {4x^2+4x+1-10x-11=0}} \right.$$\left \{ {{y=2x+1} \atop {4x^2-6x-10=0}} \right.$

4x²-6x-10=0
D=b²-4ac=36-4*4*(-10)=36-16*(-10)=36+160=$\sqrt{196}$=14

x₁=$\frac{6+14}{8} = \frac{20}{8} = 2 \frac{4}{8} =2 \frac{1}{2} =2,5$

x₂=$\frac{6-14}{8} = - \frac{8}{8} = -1$

y=2*2,5+1=6
y=2*(-1)+1=-2+1=-1

$\left \{ {{x=2,5} \atop {y=6}} \right.$

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