Simplification....at its worst.

Algebra Level 3

We all are very familiar with the simplification of a + b ± 2 a b \sqrt{a+b\pm2\sqrt{ab}} , which is a ± b \sqrt{a}\pm\sqrt{b} (for a > b a>b ). But can you simplify this? 9 77 \sqrt{9-\sqrt{77}} Express your answer in the form of a b \sqrt{a}-\sqrt{b} , where a > b a>b and enter the value of 2 a + 10 b 2a+10b .


The answer is 46.

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1 solution

Omkar Kulkarni
Mar 30, 2015

9 77 \sqrt{9-\sqrt{77}}

= 18 2 77 2 =\sqrt{\frac{18-2\sqrt{77}}{2}}

= ( 11 ) 2 2 ( 11 ) ( 7 ) + ( 7 ) 2 2 =\sqrt{\frac{\left(\sqrt{11}\right)^{2}-2\left(\sqrt{11}\right)\left(\sqrt{7}\right)+\left(\sqrt{7}\right)^{2}}{2}}

= ( 11 7 ) 2 2 =\sqrt{\frac{\left(\sqrt{11}-\sqrt{7}\right)^{2}}{2}}

= 11 2 7 2 =\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}

Cool and quick solution! My brute force approach (summarized below) was much more tedious. I should really look for shortcuts more.

9 77 = a b \sqrt{9-\sqrt{77}}=\sqrt{a}-\sqrt{b} , then 9 77 = a + b 2 a b 9-\sqrt{77}=a+b-2\sqrt{ab} .

a + b = 9 a+b=9 and 2 a b = 77 -2\sqrt{ab}=-\sqrt{77} \Rightarrow a b = 77 4 ab=\frac{77}{4} .

a 2 9 a + 77 4 = 0 a^2-9a+\frac{77}{4}=0 \Rightarrow a or b = 11 2 a \text{ or } b=\frac{11}{2} , 7 2 \frac{7}{2} .

2 a + 10 b = 2 11 2 + 10 7 2 = 11 + 35 = 46 \Rightarrow 2a+10b=2\cdot\frac{11}{2}+10\cdot\frac{7}{2}=11+35=\boxed{46} .

Alex Delhumeau - 6 years ago

Why is it : Simplification at it's worst ? Isn't this question easy ?

A Former Brilliant Member - 6 years, 2 months ago

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Idk, I took a loooooong time to think of an answer for this one :P

Omkar Kulkarni - 6 years, 2 months ago

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Okay . Are you planning on posting some Trigonometry problems ? Also are you going to participate in OPC 3 ?

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member Trigonometry occasionally, maybe, but not continuously. What's OPC 3? Never heard of it.

Omkar Kulkarni - 6 years, 2 months ago

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@Omkar Kulkarni See this .

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member Ohh. I might! I'm not so good at geometry, and especially not good at proofs. But thanks for notifying!

Omkar Kulkarni - 6 years, 2 months ago

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@Omkar Kulkarni What's the harm in trying ? It's not like everyone will see your solution , only Rajdeep will see it .

Btw , it's alright . Do take part in OPC 4 !

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member Yup! I will try, of course

Omkar Kulkarni - 6 years, 2 months ago

Did it the same way..Cool :)

Hrishik Mukherjee - 6 years, 2 months ago

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