Solve $2x^2+3x+2 \leq 4x+3$.
$x \leq -\frac{1}{2}$ and $x \geq 1$
$x \leq -1$ and $x \geq \frac{1}{2}$
$-1 \leq x\leq \frac{1}{2}$
$-\frac{1}{2} \leq x \leq 1$
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.
Wrong: Note the sign chart.
Try again.
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.
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.