Does $(x-1)^{2}+2(x+1)=0$ have a real root? Justify your answer.
(NO) To determine if the equation has real roots,we first simplify it:
$(x-1)^{2}+2(x+1)=0$
$(x^{2}-2x+1)+2x+2=0$
$x^{2}+3=0$
Now,we compare this with the standard quadratic form $ax^{2}+bx+c=0$,where $a=1, b=0, c=3$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (0)^{2}-4(1)(3) = -12$.
Since the discriminant $D < 0$,the equation has no real roots.
$(x-1)^{2}+2(x+1)=0$
$(x^{2}-2x+1)+2x+2=0$
$x^{2}+3=0$
Now,we compare this with the standard quadratic form $ax^{2}+bx+c=0$,where $a=1, b=0, c=3$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (0)^{2}-4(1)(3) = -12$.
Since the discriminant $D < 0$,the equation has no real roots.
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