In solving $|x^2-2|<-x$ by cases, why is only $x=-2$ (not $x=1$ ) accepted from the case $x^2-2\ge0$ , and only $x=-1$ (not $x=2$ ) accepted from the case $x^2-2<0$ ?

ABecause the original inequality is only ever true for negative values of

x

, by definition.

BBecause

x=-2

and

x=-1

are always the two smallest roots, regardless of which case produced them.

CBecause each case's algebraic solution must also satisfy that case's own defining condition on

x

, and only one root from each case does so.

DBecause

x=1

and

x=2

are not real numbers, so they cannot be solutions to any inequality.

Explanation: Each case's quadratic equation gives two roots, but only a root that also satisfies that case's own condition on

x

(e.g.

x^2-2\ge0

, meaning

x\le-\sqrt2

x\ge\sqrt2

) is a genuine solution of the original modulus inequality in that region; the other root is extraneous to that case.

Derived from ZIMSEC Pure Mathematics Paper 2, November 2025, Q1

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