Question

how to find the ranges of value of x for which x^2-4x+2 lies between 1 and -1

Answer 1

\(-1 < x^2-4x + 2 < 1\)

\(-1 < (x-2)^2 - 2 < 1\)

\(1 < (x-2)^2 < 3\)

Answer 2

\((x-2)^2 > 1\) AND \((x-2)^2 < 3\)

\(x-2 > 1\) OR \(x-2 < -1\) AND \((x-2) < \sqrt{3} \) OR \(x-2 > -\sqrt{3}\)

\(x > 3\) OR \(x < 1\) AND \(x < 2+\sqrt{3} \) OR \(x > 2-\sqrt{3}\)

that gives : x must lie in \( (3, 2+\sqrt{3}) \cup (1, 2-\sqrt{3})\)