Exercise 4 | Wiskunde op Tilburg University

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

$x\neq 1$ .

Antwoord 2 feedback

Wrong:

$(1^2+1-2)(1-3)=0 \ngtr 0= -6(1-1)$ .

Try again.

Antwoord 3 feedback

Wrong:

$(2^2+2-2)(2-3)=-4> -6 = -6(2-1)$ .

Try again.

Antwoord 4 feedback

Wrong:

$((\frac{1}{2})^2+\frac{1}{2}-2)(\frac{1}{2}-3)=3\frac{1}{8} > 3 = -6(\frac{1}{2}-1)$ .

Try again.