# Absolute Value Inequalities rule

Solve the inequality and write the solution in brackets

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

What is the rule here?

• do i have to separate the inequality into two:

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

and

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

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

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