Csdrhrt

$a < 0 < b$ 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?

$(ax -b)(bx-a)ge 0$ means either $ax-b ge 0; bx-a ge 0$ or $ax-ble 0; bx-ale 0$.

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.

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
$$xinbiggl[frac ba,frac abbiggr].$$

You can rewrite the inequality as
$$
ableft(x-frac{b}{a}right)left(x-frac{a}{b}right)ge0
$$
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
$$
left[frac{b}{a},frac{a}{b}right]
$$

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

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

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 $ygeq0$ precisely when $frac{a}{b}leq xleqfrac{b}{a}$.

We must have that either $$ax-bge 0\bx-age 0$$or$$ax-ble 0\bx-ale 0$$In the first case we have $$xle {bover a},xge {aover b}$$which is impossible since ${bover a}<{aover b}$. For the latter case we obtain$$xge {bover a},xle {aover b}$$therefore the solution would become$${bover a}le xle {aover b}$$