Question

# $$|x-3|^{3x^2 – 10x + 3} = 1$$. Find $$x$$.

Solution:

There can be three cases:

CASE I : $$3x^2 – 10x + 3 = 0$$ and $$|x-3| \neq 0$$

$$3x^2 – 10x + 3 = 0$$ and $$x-3 \neq 0$$
$$\Rightarrow 3x^2 – 9x – x + 3 = 0$$ and $$x \neq 3$$
$$\Rightarrow 3x(x – 3) – 1(x – 3) = 0$$ and $$x \neq 3$$
$$\Rightarrow (3x – 1)(x – 3) = 0$$ and $$x \neq 3$$
$$x = \frac{1}{3}$$ or $$x = 3$$ and $$x \neq 3$$

Therefore from CASE I , $$x = \frac{1}{3}$$

CASE II : $$|x – 3| = 1$$

$$x\;-\;3 \pm 1$$
$$\Rightarrow x = 4$$ OR $$x = 2$$

From CASE II , $$x = 4 , x = 2$$

CASE III : $$|x – 3| = -1$$ and $$3x^2 – 10x + 3$$ is an even number or is equal to 0

But $$|x – 3| = -1$$ is not possible.

Hence we get $$x \in \{\frac{1}{3} , 4 , 2 \}$$

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