1.

Simplify:|x - 3| + 2|x + 1| = 4.

Answer»

|x - 3| + |x + 1| = 4

⇒ |x - 3| + 2|x + 1| - 4 = 0 f(x) (Let)

f(x) = \(\begin{cases}-(x-3)-2(x+1)-4&;x\leq-1\\-(x-3)+2(x+1)-4&;-1\leq x\leq 3\\ x-3+2(x+1)-4&;x\geq3\end{cases}\)

 = \(\begin{cases}-3x-3&;x\leq-1\\x+1&;-1\leq x\leq3\\3x-6&;x\geq3\end{cases}\)

 ∵ -3x - 3 = 0 ⇒ x = -1 which satisfies x \(\leq-1\)

x + 1 = 0 ⇒ x = -1 which satisfies -1 \(\leq x\leq3\)

3x - 6 = 0 ⇒ x = 2 which does not satisfy x \(\geq\) 3.

∴ x = -1 is only zero of f(x).

Hence,  |x - 3| + 2 |x + 1| = 4 satisfies by x = -1



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