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 $-1<\beta<1\frac{1}{2}$.
$D(\beta)=(4\beta)^2-4\cdot 2\cdot (\beta+3)=16\beta^2-8\beta-24$.
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 $\beta$.
See Extra explanation: zeros or Example 3 (film).
See Extra explanation: zeros or Example 3 (film).
Antwoord 3 feedback
Wrong: $(4\beta)^2\neq 4\beta^2$.
Try again.
Try again.
Antwoord 4 feedback