Austrailan Mathematics Competition Warmup

Algebra Level 1

Which of the following is equal to 1 2 5 − 3 \frac { 1 }{ 2\sqrt { 5 } -\sqrt { 3 } } ?

1) 2 5 − 3 7 \frac { 2\sqrt { 5 } -\sqrt { 3 } }{ 7 }

2) 2 5 + 3 17 \frac { 2\sqrt { 5 } +\sqrt { 3 } }{ 17 }

3) 2 5 − 3 17 \frac { 2\sqrt { 5 } -\sqrt { 3 } }{ 17 }

4) 2 5 + 3 7 \frac { 2\sqrt { 5 } +\sqrt { 3 } }{ 7 }

3 1 4 2

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Isaiah Simeone by isaiah simeone , 23, Australia

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5 solutions

1 2 5 − 3 × 2 5 + 3 2 5 + 3 = 2 5 + 3 20 − 3 = 2 5 + 3 17 \color{#3D99F6}{\frac{1}{2\sqrt5-\sqrt3}\times\frac{2\sqrt{5}+\sqrt{3}}{2\sqrt{5}+\sqrt{3}}\\ =\frac{2\sqrt{5}+\sqrt{3}}{20-3}=\frac{2\sqrt{5}+\sqrt{3}}{17}}

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Harihar Singh
Dec 9, 2014

we multiply by 2root5+root3 in numerator and denomirator and we get the final result

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Ashutosh Tiwari
Dec 4, 2014

A question which can be solved mentally. Very easy question . Multiplying and dividing by conjugate. In denominator we get (a-b)*(a+b)i.e is a^ - b^2 .And on numerator we get the conjugate of denominator.

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Sanjeet Raria
Sep 25, 2014

1 2 √ 5 − √ 3 2 √ 5 + √ 3 2 √ 5 + √ 3 = 2 √ 5 + √ 3 20 − 3 \frac{1}{2√5-√3}\frac{2√5+√3}{2√5+√3}=\frac{2√5+√3}{20-3} = 2 √ 5 + √ 3 17 =\frac{2√5+√3}{17}

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Hadia Qadir
Aug 30, 2015

2 because we multiply the denominator and numerator by (2 *root of 5) +root of 3

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