How can I use quadratic formula with an inequation like this $(x-3)(x^2-5x+5)\le0$

Question

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

$(x-3)(x^2-5x+5)\le0$

Answer 1

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.

Answer 2

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 $x<a$

Answer 3

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:)