Animated Solution for Mathematics - Quadratic Equations: Find all real values of $x$ which satisfy $x^2-3x+2>0$ and $x^2-2x-4\le 0$

We begin with the first quadratic inequality: $x^2-3x+2>0$. This represents a parabola that opens upward since the coefficient of $x^2$ is positive.

To find where the parabola rises above the $x$-axis, we factorize the expression. We seek two numbers that multiply to $2$ and add to $-3$, which are $-1$ and $-2$.

This gives us the factored form:

The roots of this equation are $x=1$ and $x=2$. Because the parabola opens upward, it remains above the $x$-axis outside of these roots.

Thus, our first solution set is:

Next, we address the second inequality: $x^2-2x-4\le 0$. Since this does not factorize cleanly, we apply the quadratic formula:

Substituting $a=1$, $b=-2$, and $c=-4$, we obtain:

Because we require the region where the expression is less than or equal to zero, we are interested in the "valley" of the parabola, which lies between the roots.

Therefore, our second solution set is:

Finding the Intersection

To solve the system, we must find the overlap between our two solution sets. We compare the intervals $(-\infty ,1)\cup (2,\infty )$ and $[1-\sqrt{5},1+\sqrt{5}]$.

Note that $1-\sqrt{5}\approx -1.23$ and $1+\sqrt{5}\approx 3.23$. By observing the number line, we identify the regions where both conditions are satisfied simultaneously.

The intersection of these two sets is:

$$x\in [1-\sqrt{5},1)\cup (2,1+\sqrt{5}]$$

You have successfully navigated the constraints and identified the precise range of values that satisfy both inequalities. Keep this analytical mindset, and no problem will ever be too complex for you.