Exercise 11 | Wiskunde op Tilburg University

Consider

$y(x)=2x^2+4\beta x +\beta +3$ . Determine all the values of

$\beta$ such that the graph of

$y(x)$ does not intersect the

$x$ -axis.

Antwoord 1 correct

Correct

Antwoord 2 optie

All

$\beta$

Antwoord 2 correct

Fout

Antwoord 3 optie

$\frac{1}{4}-\frac{1}{4}\sqrt{7} <\beta <\frac{1}{4}+\frac{1}{4}\sqrt{7}$

Antwoord 3 correct

Fout

Antwoord 4 optie

The correct answer is not among the other options.

Antwoord 4 correct

Fout

Antwoord 1 optie

$-1<\beta<1\frac{1}{2}$

Antwoord 1 feedback

Correct:

$y(x)$ does not intersect the

$x$ -axis when the discriminant

$D<0$ .

$D(\beta)=(4\beta)^2-4\cdot 2\cdot (\beta+3)=16\beta^2-8\beta-24$ .

We determine the zeros of

$D(\beta)$ .

$\beta_1=\dfrac{8-\sqrt{ (-8)^2-4\cdot 16\cdot -24}}{2\cdot 16}=-1$ and

$\beta_2=\dfrac{8+\sqrt{(-8)^2-4\cdot 16\cdot -24}}{2\cdot 16}=1\frac{1}{2}$ .

Via a sign chart (with for instance

$D(-2)=56$ ,

$D(0)=-24$ and

$D(2)=24$ ) we find

$D(\beta)<0$ if

$-1<\beta<1\frac{1}{2}$ .

Antwoord 2 feedback

Wrong: The discriminant of

$y(x)$ depends on

Antwoord 3 feedback

Wrong:

$(4\beta)^2\neq 4\beta^2$ .

Try again.

Antwoord 4 feedback

Wrong: The correct answer is among them.

See Extra explanation: zeros or Example 3 (film).