Exercise 12 | Wiskunde op Tilburg University

Solve $2x^2+3x+2 \leq 4x+3$ .

Antwoord 1 correct

Correct

Antwoord 2 optie

$x \leq -\frac{1}{2}$ and $x \geq 1$

Antwoord 2 correct

Fout

Antwoord 3 optie

$x \leq -1$ and $x \geq \frac{1}{2}$

Antwoord 3 correct

Fout

Antwoord 4 optie

$-1 \leq x\leq \frac{1}{2}$

Antwoord 4 correct

Fout

Antwoord 1 optie

$-\frac{1}{2} \leq x \leq 1$

Antwoord 1 feedback

Correct: $2x^2+3x+2 = 4x+3 \Leftrightarrow 2x^2-x-1=0$ .

We define $f(x)=2x^2-x-1$ and determine $f(x)=0$ .

$x_1=\dfrac{1-\sqrt{(-1)^2-4\cdot 2\cdot -1}}{2\cdot 2}=-\frac{1}{2}$ or $x_2=\dfrac{1+\sqrt{(-1)^2-4\cdot 2\cdot -1}}{2\cdot 2}=1$ .

Via a sign chart (with for instance $f(-1)=2$ , $f(0)=-1$ and $f(2)=5$ ) we find $-\frac{1}{2}\leq x \leq 1$ .

Go on.

Antwoord 2 feedback

Wrong: Note the sign chart.

Try again.

Antwoord 3 feedback

Wrong: Note that the solutions of the quadratic equations are given by
$x=\frac{-b-\sqrt{b^2-4ac}}{2a} \text{ and } x=\frac{-b+\sqrt{b^2-4ac}}{2a}.$ .

See Extra explanation: zeros.

Antwoord 4 feedback

Wrong: Note that the solutions of the quadratic equations are given by
$x=\frac{-b-\sqrt{b^2-4ac}}{2a} \text{ and } x=\frac{-b+\sqrt{b^2-4ac}}{2a}.$ .

See Extra explanation: zeros.