Evaluate after rationalising the denominator of (25/√40-√80), it being given that √5=2.236 and √10=3.162.

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Solution

25/√40-√80
=25/2√10-4√5
=25/2(√10-2√5)
=25(√10+2√5)/2(√10-2√5)(√10+2√5)
=25(√10+2√5)/2(10-20)
=-25(√10+2√5)/20
=-5(√10+2√5)/4
=-5(3.162+2×2.236)/4
=-5(3.162+4.472)/4
=-5×7.634/4
=-5×1.9085
=-9.5425

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Similar questions

\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}

\sqrt{5} = 2.236

\sqrt{10} = 3.162

\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}

\sqrt{5} = 2.236 and \sqrt{10} = 3.162

\begin{array}{rcl} \frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}} & = & \frac{15}{\sqrt{10} + 2 \sqrt{5} + 2 \sqrt{10} - \sqrt{5} - 4 \sqrt{5}} \\ = & \frac{15}{3 \sqrt{10} - 3 \sqrt{5}} = \frac{5}{\sqrt{10} - \sqrt{5}} \end{array}

Evaluate $\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$ , it us given that $\sqrt{5} = 2.236$ and $\sqrt{10} = 3.162$

\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} + \sqrt{5} - \sqrt{80}}

\sqrt{5} = 2.236 a n d \sqrt{10} = 3.162

\sqrt{2} = 1.414, \sqrt{3} = 1.732

\sqrt{5} = 2.236

\frac{\sqrt{10} - \sqrt{5}}{2}

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