Given that aa is a positive constant, solve the inequality ∣x−3a∣>∣x−a∣|x-3a|>|x-a|.

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Worked solution

Q4

1. Eliminate the modulus by squaring both sides of the inequality.

∣x−3a∣>∣x−a∣|x-3a| > |x-a|
(x−3a)2>(x−a)2(x-3a)^2 > (x-a)^2

2. Expand the squared expressions.

x2−6ax+9a2>x2−2ax+a2x^2 - 6ax + 9a^2 > x^2 - 2ax + a^2

3. Simplify the inequality by grouping like terms.

9a2−a2>x2−x2−2ax+6ax9a^2 - a^2 > x^2 - x^2 - 2ax + 6ax
8a2>4ax8a^2 > 4ax
2a2>ax2a^2 > ax

4. Solve for $x$, using the given condition that $a$ is positive.

Since aa is a positive constant (a>0a > 0), dividing both sides by aa preserves the inequality sign:
2a>x2a > x
x<2ax < 2a

Answer

x<2ax < 2a

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