How do we find the solution root of this inequality?

Question

$a < 0 < b$ and $|a| < b$ $$(ax – b)(bx – a) ≥ 0$$

How do we find the solution root of this inequality?

My attempt:

$$(ax – b)(bx – a) = a^2b^2x^2-a^2x-b^2x+ab $$

$$a^2b^2x^2-a^2x-b^2x+ab ≥ 0 $$

$$a^2b^2x^2+ab ≥ a^2x+b^2x$$

I, however, could not proceed from there. Perhaps I made it take longer than it actually does. Could you assist?

Answer 1

$(ax -b)(bx-a)\ge 0$ means either $ax-b \ge 0; bx-a \ge 0$ or $ax-b\le 0; bx-a\le 0$.

So Case 1: $ax -b \ge 0$ and $bx - a \ge 0$

Then $ax \ge b$ so $x \le \frac ba$ ($a <0$).

Also $bx \ge a$ so $x \ge \frac ab$ ($b > 0$).

So $\frac ab \le x \le \frac ba $

This may or may not be possible depending upon whether $\frac ab < \frac ba; \frac ab = \frac ba; $ or $\frac ab > \frac ba$.

We have $a < 0 < -a=|a| < b$ so $\frac ab < -\frac ab < 1$ and $-1 < \frac ab < 0$.

And have $a < 0 < -a < b$ so $1 > 0 > -1 > \frac ba$.

So $\frac ba < \frac ab < 0$ and this is a contradiction.

.....

And Case 2: $ax -b \le 0$ and $bx - a \le 0$

Then $ax \le b$ so $x \ge \frac ba$.

Also $bx \le a$ so $x \le \frac ab$.

So $\frac ba \le x \le \frac ab $

and that is consistant as $\frac ba < \frac ab$.

....

So the solution set is $\frac ba \le x \le \frac ab$. (Note: $x$ is definitely negative.

Answer 2

First, you have that $(ax-b)(bx-a)=abx^2-(a^2+b^2)x+ab$. Hence, $$abx^2-(a^2+b^2)x+ab=0 \iff x=\frac{(a^2+b^2)\pm \sqrt{(a^2+b^2)^2-4a^2b^2}}{2ab}=\frac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}.$$ So because $|a|<b$, then $\sqrt{(a^2-b^2)^2}=|a^2-b^2|=b^2-a^2$, so $$abx^2-(a^2+b^2)x+ab=0 \iff x=\frac{a^2+b^2\pm(b^2-a^2)}{2ab}\iff x=\frac{b}{a}\ {\rm and}\ x=\frac{a}{b},$$ where $b/a<a/b<0$.

Because the 2nd grade polynomial have negative leader coeficient, we have that $$(ax-b)(bx-a)\ge0 \iff x\in\left[\frac{b}{a},\frac{a}{b}\right].$$

Answer 3

There's a theorem on the sign of quadratic polynomials on $\mathbf R$, which can summarised as follows:

A quadratic polynomial has the sign of its leading coefficient, except between its (real) roots, if any.

Here the leading coefficient is $ab$, which is negative. So this polynomial is positive between its roots $a/b$ and $b/a$. The hypothesis about $a$ and $b$ imply that $b/a<-1<a/b<0$, so the solution is $$x\in\biggl[\frac ba,\frac ab\biggr].$$

Answer 4

You can rewrite the inequality as $$ ab\left(x-\frac{b}{a}\right)\left(x-\frac{a}{b}\right)\ge0 $$ Since $a<0$ and $b>0$, it becomes $$ \left(x-\frac{b}{a}\right)\left(x-\frac{a}{b}\right)\le0 $$ which is satisfied in the closed interval delimited by the roots.

Now we just have to decide what's the larger root: $$ \frac{b}{a}<\frac{a}{b}\iff b^2>a^2 $$ (remember that $a<0$ and $b>0$). Since $b^2>a^2$ is given, we have that the solution set is $$ \left[\frac{b}{a},\frac{a}{b}\right] $$

Answer 5

The equation $y=(ax-b)(bx-a)$ defines a parabola with zeroes $\frac{a}{b}$ and $\frac{b}{a}$, where $\frac{b}{a}<\frac{a}{b}$ because $|a|<b$ and $a<0$. Its leading coefficient $ab$ is negative, so $y\geq0$ precisely when $\frac{a}{b}\leq x\leq\frac{b}{a}$.

Answer 6

We must have that either $$ax-b\ge 0\\bx-a\ge 0$$or$$ax-b\le 0\\bx-a\le 0$$In the first case we have $$x\le {b\over a},x\ge {a\over b}$$which is impossible since ${b\over a}<{a\over b}$. For the latter case we obtain$$x\ge {b\over a},x\le {a\over b}$$therefore the solution would become$${b\over a}\le x\le {a\over b}$$