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If \(15<\sqrt{m}<16\), what is the correct range for \(m\)?

Answer and explanation

Correct answer: \(225<m<256\)

Since \(\sqrt{m}\) is positive, squaring every part of the inequality preserves the order. Squaring gives \(15^2<m<16^2\), i.e. \(225<m<256\). Option C (\(196<m<225\)) would correspond to squaring \(14<\sqrt{m}<15\) and so is incorrect; option B simply restates the original range for \(\sqrt{m}\) rather than for \(m\); option D corresponds to squaring \(16<\sqrt{m}<17\). Exam tip: when all terms are nonnegative you may square an inequality termwise — then compute the numerical squares to get the correct interval for the variable inside the square.

Related tags

Real NumbersEstimationInequalitySquaresRadicalsNumber SystemsMathematicsClass 9 Mcq

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