How do I solve a quadratic inequality without factoring?

Question

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.

Answer 1

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 $$\left|x-\frac{5}{2}\right|\le \frac{1}{2}$$

Answer 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 $x\in[2,3]$

Answer 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 $2\leq x\leq3.$

Answer 4

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

Answer 5

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.

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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 $\mathbb R$.

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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,

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

Answer 6

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.