Write in ascending order:
(i) 𝟑√𝟓 𝒂𝒏𝒅 𝟒√𝟑
(ii) 𝟐 √𝟓 𝟑 𝒂𝒏𝒅 𝟑 ³√𝟐
(iii) 𝟔√𝟓, 𝟕√𝟑 𝒂𝒏𝒅 𝟖√𝟐
(i) 3 \sqrt{5}=\sqrt{3^{2} \times 5}=\sqrt{45}, 4 \sqrt{3}=\sqrt{4^{2} \times 3}=\sqrt{48} We know that, 45 < 48
\therefore \sqrt{45}<\sqrt{48} \Rightarrow 3 \sqrt{5}<4 \sqrt{3}
(ii) 2 \sqrt[3]{5}=\sqrt[3]{2^{3} \times 5}=\sqrt[3]{40}, 3 \sqrt[3]{2}=\sqrt[3]{3^{3} \times 2}=\sqrt[3]{54}
We know that,40 < 54
\begin{array}{l} \Rightarrow \sqrt[3]{40}<\sqrt[3]{54} \\ \Rightarrow 2 \sqrt[3]{5}<3 \sqrt[3]{2} \end{array}
(iii) 6 \sqrt{5}=\sqrt{6^{2} \times 5}=\sqrt{180}
We know that, 128 < 147 < 180
\begin{array}{l} \therefore \sqrt{128}<\sqrt{147}<\sqrt{180} \\ \Rightarrow 8 \sqrt{2}<7 \sqrt{3}<6 \sqrt{5} \end{array}
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