Absolute Value Inequalities rule

Question

Solve the inequality and write the solution in brackets

$$\left|6-4x\right| \geq \left|x-2\right|$$

What is the rule here?

$$\left|6-4x\right| \geq 0$$

and

$$\left|x-2\right| \geq 0$$

Please need an explanation

lab bhattacharjee · Accepted Answer · 2015-09-18 10:40:44Z

0

HINT: $$|p+iq|\ge |a+ib|\iff\sqrt{p^2+q^2}\ge\sqrt{a^2+b^2} \iff p^2+q^2\ge a^2+b^2$$

If $q=b=0,$ $$|p|\ge |a|\iff p^2\ge a^2$$

answered Sep 18 '15 at 10:25

lab bhattacharjee

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$\begingroup$ Basically like this: 6^2 - 4x^2 ≥ x^2 -2^2 $\endgroup$ – question Sep 18 '15 at 10:40
$\begingroup$ @question, No $$(6-4x)^2\ne6^2-(4x)^2$$ in general $\endgroup$ – lab bhattacharjee Sep 18 '15 at 10:41
$\begingroup$ @question $(6-4x)^2=6^2-2\cdot 6\cdot 4x+(4x)^2$. Use $(a-b)^2=a^2-2ab+b^2$. $\endgroup$ – user236182 Sep 18 '15 at 10:43

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