Solve $(x^2+x-2)(x-3) > -6(x-1)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x>0$
Antwoord 2 correct
Fout
Antwoord 3 correct
Fout
Antwoord 4 optie
$x>1$
Antwoord 4 correct
Fout
Antwoord 1 optie
The correct answers is not among the other options.
Antwoord 1 feedback
Correct: $$\begin{align*}
(x^2+x-2)(x-3) = -6(x-1) & \Leftrightarrow (x-1)(x+2)(x-3) =-6(x-1)\\
& \Leftrightarrow (x-1)(x^2-x-6)+6(x-1) = 0\\
& \Leftrightarrow (x-1)(x^2-x) = 0\\
& \Leftrightarrow x(x-1)^2 = 0\\
& \Leftrightarrow x = 1 \mbox{ or } x=0.
\end{align*}$$
Since $(x-1)^2>0$ if $x\neq 1$, it must hold that $x>0$ and $x\neq 1$.
(x^2+x-2)(x-3) = -6(x-1) & \Leftrightarrow (x-1)(x+2)(x-3) =-6(x-1)\\
& \Leftrightarrow (x-1)(x^2-x-6)+6(x-1) = 0\\
& \Leftrightarrow (x-1)(x^2-x) = 0\\
& \Leftrightarrow x(x-1)^2 = 0\\
& \Leftrightarrow x = 1 \mbox{ or } x=0.
\end{align*}$$
Since $(x-1)^2>0$ if $x\neq 1$, it must hold that $x>0$ and $x\neq 1$.
Antwoord 2 feedback
Wrong: $(1^2+1-2)(1-3)=0 \ngtr 0= -6(1-1)$.
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Try again.
Antwoord 3 feedback
Wrong: $(2^2+2-2)(2-3)=-4> -6 = -6(2-1)$.
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Try again.
Antwoord 4 feedback
Wrong: $((\frac{1}{2})^2+\frac{1}{2}-2)(\frac{1}{2}-3)=3\frac{1}{8} > 3 = -6(\frac{1}{2}-1)$.
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Try again.