State whether the quadratic equation $x(1-x)-2=0$ has two distinct real roots. Justify your answer.
(NO) Given equation is $x(1-x)-2=0$.
Expanding the terms,we get $x-x^{2}-2=0$.
Rearranging the equation in the standard form $ax^{2}+bx+c=0$,we get $x^{2}-x+2=0$.
Comparing this with $ax^{2}+bx+c=0$,we have $a=1, b=-1, c=2$.
The discriminant $D$ is given by $D=b^{2}-4ac$.
Substituting the values,$D=(-1)^{2}-4(1)(2) = 1-8 = -7$.
Since $D < 0$,the quadratic equation $x^{2}-x+2=0$ has no real roots (it has two distinct imaginary roots).
Expanding the terms,we get $x-x^{2}-2=0$.
Rearranging the equation in the standard form $ax^{2}+bx+c=0$,we get $x^{2}-x+2=0$.
Comparing this with $ax^{2}+bx+c=0$,we have $a=1, b=-1, c=2$.
The discriminant $D$ is given by $D=b^{2}-4ac$.
Substituting the values,$D=(-1)^{2}-4(1)(2) = 1-8 = -7$.
Since $D < 0$,the quadratic equation $x^{2}-x+2=0$ has no real roots (it has two distinct imaginary roots).
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