Question

a) {x∈ N | x²-4x+3=0} = {1,3} ;
b) {x∈ N | x²+x-2=0} = {1}
Help Va rog :o3

Answer 1

a) $x^{2} -4x+3=0$

Δ=$b^{2}-4ac=(- 4)^{2}-4*1*3=16-12=4$

$x_{1} = \frac{-b+ \sqrt{delta} }{2a} = \frac{-(-4)+ \sqrt{4} }{2*1} = \frac{4+2}{2} =\frac{6}{2}=3$

$x_{2} = \frac{-b- \sqrt{delta} }{2a} = \frac{-(-4)- \sqrt{4} }{2*1} = \frac{4-2}{2} =\frac{2}{2}=1$

x∈{1; 3}

b) $x^{2} +x-2=0$

Δ=$b^{2}-4ac=1^{2}-4*1*(-2)=1-(-8)=1+8=9$

$x_{1} = \frac{-b+ \sqrt{delta} }{2a} = \frac{-(-1+ \sqrt{9} }{2*1} = \frac{-1+3}{2} =\frac{2}{2}=1$

$x_{2} = \frac{-b- \sqrt{delta} }{2a} = \frac{-1- \sqrt{9} }{2*1} = \frac{-1-3}{2} =\frac{-4}{2}=-2$

x∈{-2;1}