How can I use quadratic formula with an inequation like this $(x-3)(x^2-5x+5)\le0$ I know how to use the quadratic formula with an inequation.
How can I make use the quadratic formula?
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
With an inequation like this
$(x-3)(x^2-5x+5)\le0$
 A: Firstly, inequality. Please try to be accurate with terminology.
For the question: First, you want to find the critical values, that is, the $x$ where $(x-3)(x^2-5x+5)=0$
The first solution $x=3$ is trivial, then use the quadratic formula on $x^2-5x+5$ to find the other two.
I'll call them $x_1$ and $x_2$ (you should work them out) and note that $x_1<3<x_2$.
We now have four different intervals to test. Notice that between two adjacent zeroes of a function, the function is always positive or always negative. Hence if we pick an arbitrary $x$ in each of the intervals:
$$x<x_1$$
$$x_1<x<3$$
$$3<x<x_2$$
$$x>x_2$$
And see if the result is positive or negative when we put it in our function. If its negative, the interval it is in fits the $\leq 0$ criterion.
A: Using the formula we have $$f(x)=(x-a)(x-3)(x-b)\le0$$
where $a=\dfrac{5-\sqrt5}2<3<b=\dfrac{5+\sqrt5}2$
For $f(x)=0,x=a,b$ or $3$
For $f(x)<0,$ odd number of multiplicands need to be $<0$
which will occur if $3<x<b$ or if $x<a$
A: Let $f(x)=x^2-5x+5$, we see that here the coefficient of $x^2$ term is positive and $f(3)=-1<0$, so 3 lies between the roots of $f(x)$. Let the smaller root be $\alpha$. So we have,
$$(x-\alpha)(x-3)(x-\beta)\leq 0$$
Determine the sign on each of the intervals $(-\infty,\alpha), (\alpha,3), (3,\beta)$ and $(\beta,\infty)$. Therefore, the required solution is $$x\in (-\infty,\alpha]\cup [3,\beta]$$ where the values of $\alpha$ and $\beta$ are to be found by the quadratic formula.
Hope it helps:)
