State whether the quadratic equation $(x+1)(x-2)+x=0$ has two distinct real roots. Justify your answer.
(A) Given equation is $(x+1)(x-2)+x=0$.
Expanding the terms:
$x^2 - 2x + x - 2 + x = 0$
Simplifying the equation:
$x^2 - 2 = 0$
Comparing this with the standard form $ax^2 + bx + c = 0$,we get:
$a = 1, b = 0, c = -2$
The discriminant $D$ is given by $D = b^2 - 4ac$.
Substituting the values:
$D = (0)^2 - 4(1)(-2) = 0 + 8 = 8$.
Since $D > 0$,the quadratic equation has two distinct real roots.
Expanding the terms:
$x^2 - 2x + x - 2 + x = 0$
Simplifying the equation:
$x^2 - 2 = 0$
Comparing this with the standard form $ax^2 + bx + c = 0$,we get:
$a = 1, b = 0, c = -2$
The discriminant $D$ is given by $D = b^2 - 4ac$.
Substituting the values:
$D = (0)^2 - 4(1)(-2) = 0 + 8 = 8$.
Since $D > 0$,the quadratic equation has two distinct real roots.
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