# How do I solve a quadratic inequality without factoring?

Let's say I've got the formula $$-x^{2}+5x-6\geq 0$$

How should I go about solving this without factoring? I'm okay with using the quadratic formula, but I don't want to factor it.

• Go for square completion. But what do you have against factoring ? – Yves Daoust Jul 17 at 13:38
• Why wouldn't you want to factor it? – Simply Beautiful Art Jul 17 at 13:39
• @Simply Beautiful Art Because he has a solution by factoring. – Michael Rozenberg Jul 17 at 13:46
• @MichaelRozenberg And where does it say that in the post? – Simply Beautiful Art Jul 17 at 13:54
• @MichaelRozenberg: presumably because he thinks that factoring is harder. (Which might be true if done by trial or error or other empirical methods, but the blame is not to be put on factoring per se.) – Yves Daoust Jul 17 at 13:58

Multiplying by $$-1$$ we get $$x^2-5x+6\le 0$$ completing the square $$x^2-5x+\frac{25}{4}-\frac{25}{4}+\frac{24}{4}\le 0$$ Can you finish? And then you will get $$\left|x-\frac{5}{2}\right|\le \frac{1}{2}$$

We know that $$f(2)=f(3)=0$$ Also $$f'(x)\ge0$$ for $$x\le2$$ and $$f'(x)\le0$$ for $$x\ge3$$ so the solution is $$x\in[2,3]$$

$$-x^2+5x-6=ax^2+bx+c.$$ We have: $$a=-1$$, $$b=5$$ and $$c=-6$$.

$$\Delta=b^2-4ac=25-4(-1)(-6)=1=1^2.$$ Thus, for roots of $$-x^2+5x-6$$ we obtain: $$x_1=\frac{-5+1}{2(-1)}=2$$ and $$x_2=\frac{-5-1}{2(-1)}=3.$$

Now, draw a parabola $$y=-x^2+5x-6.$$

We see that $$-x^2+5x-6\geq0$$, id est, $$y\geq0$$ for $$2\leq x\leq3.$$

If you are OK with the quadratic formula you can find the zeros of your quadratic equation.

Then test the intervals $$(-\infty , x_1)$$, $$(x_1,x_2)$$,and $$(x_2, \infty)$$ to see which interval satisfies your inequality.

In case that your roots are complex you just test one arbitrary number to see if the whole real line satisfies the inequality or you have no solution.

In your case the roots are $$2$$ and $$3$$ so you test the intervals $$(-\infty ,2)$$, $$(2,3)$$, and $$(3,\infty)$$

The interval $$[2,3]$$ satisfies the inequality.

You can't !

Because the solution of the inequation will be one of

$$x\in[a,b]$$ or $$x\notin[a,b]$$

where $$a,b$$ are the roots, which you need to provide, and the factorization is

$$(x-a)(x-b).$$

So solving and factoring are essentially the same operation.

Even when the roots are complex, you have to compute the discriminant and you can conclude simultaneously that there are no roots and no factorization in $$\mathbb R$$.

By completing the square, you can reach a form like

$$(x-p)^2\le q$$ and claim that this is the solution. But you are one micron from the factorization, by the difference of two squares,

$$((x-p)-\sqrt q)((x-p)+\sqrt q)\le0.$$

Edit : seems like im not paying attention so well to the question

There's plenty way to turn quadratic into linear multiplication, by quadratic formula, complete the square, grouping term