Roots of Roots

Algebra Level 2

The value of 16 + 240 \sqrt{16+\sqrt{240}} can be expressed in the form a + b \sqrt{a} + \sqrt{b} , where a a and b b are integers. Find a + b a+b .


The answer is 16.

Bufang Liang by Bufang Liang , 29, USA

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2 solutions

Isaac Buckley
Aug 27, 2015

16 + 240 = 10 + 2 10 × 6 + 6 = ( 10 + 6 ) 2 = 10 + 6 \sqrt{16+\sqrt{240}}=\sqrt{10+2\sqrt{10\times 6}+6}=\sqrt{(\sqrt{10}+\sqrt{6})^2}=\sqrt{10}+\sqrt{6}

∴ a + b = 10 + 6 = 16 \large \therefore a+b=10+6=\boxed{16}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

That was easy

Sai Ram - 5 years, 9 months ago
Bufang Liang
Aug 27, 2015

16 + 240 = a + b 16 + 240 = ( a + b ) 2 = a + 2 a b + b 16 = a + b 240 = 2 a b a = 10 , b = 6 \sqrt{16+\sqrt{240}} = \sqrt{a} + \sqrt{b} \\ 16+\sqrt{240} = (\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b \\ 16 = a + b \\ 240 = 2\sqrt{ab} \\ a = 10, b = 6

0 Helpful 0 Interesting 0 Brilliant 0 Confused

We had to find a+b so no need to find individual values of a and b.

Kushagra Sahni - 5 years, 9 months ago

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