Will you square ???

Algebra Level 4

If $\sqrt { \sqrt { 3-\sqrt { 8 } } +\sqrt { 6+\sqrt { 32 } } } =\sqrt { \sqrt { a+b\sqrt { c } } }$ ,where $a,b,c\in$ natural numbers and $c$ is square free.Find $a+b+c$ .

The answer is 15.

4 solutions

Shivamani Patil
Aug 26, 2014

First notice that $\sqrt { 3-\sqrt { 8 } } =\sqrt { 2-2\sqrt { 2 } +1 } =\sqrt { { (\sqrt { 2 } -1) }^{ 2 } } =\sqrt { 2 } -1$

Also

$\sqrt { 6+\sqrt { 32 } } =\sqrt { 2+2\times 2\sqrt { 2 } +4 } =\sqrt { { (\sqrt { 2 } +2) }^{ 2 } } =\sqrt { 2 } +2$

Therefore

$\sqrt { \sqrt { 3-\sqrt { 8 } } +\sqrt { 6+\sqrt { 32 } } }=\sqrt { \sqrt { 2 } -1+\sqrt { 2 } +2 } =\sqrt { 2\sqrt { 2 } +1 }$

Now,

$\sqrt { 2\sqrt { 2 } +1 } =\sqrt { \sqrt { { (2\sqrt { 2 } +1) }^{ 2 } } }$

$\sqrt { \sqrt { 8+2\times 2\sqrt { 2 } +1 } } =\sqrt { \sqrt { 9+2\times 2\sqrt { 2 } } } =\sqrt { 9+4\sqrt { 2 } }$

Now by comparing RHS of original equation we get $a=9,b=4,c=2$

And $a+b+c=15$

i just wonder is it true if i write :

$\sqrt{9+4\sqrt{2}}$ = $\sqrt{9+2\sqrt{8}}$ = $\sqrt{9+1\sqrt{32}}$

the question doesn't say that $\sqrt{c}$ must be in simplest radical form, so in that case, you can find 3 different answers for this question.

Fahrul Razi - 6 years, 9 months ago

Actually, it does say that c is square-free

Abhishek Gorai - 6 years, 9 months ago

@shivamani patil There's a typo in the $1^{st}$ line.It should be $\sqrt{2}-1$ ...

Satvik Golechha - 6 years, 9 months ago

Yeah... True!

Krishna Ar - 6 years, 9 months ago

You should specify that $a,b,c$ are natural numbers, not real numbers and that $c$ is square free. Otherwise, there are multiple answers for the question. @shivamani patil

Nanayaranaraknas Vahdam - 6 years, 9 months ago

All that fixed

shivamani patil - 6 years, 9 months ago

integrate cosx/x from npi to (n+1)pi

Neeraj Gupta - 6 years, 9 months ago

i have solved it by comparing values in their radicals 3+6=9 at L.H.S which equals toa 'a' at R.H.S then common in 8 &32 is 2 which is equal to ' b' then multipy the remainder in 8 ( 2 )and 32 is (8 ) which is 16 sq.root of 16 is 4 =c so a+b+c=9+2+4=15

esha aslam - 6 years, 8 months ago

Chew-Seong Cheong
Aug 26, 2014

It is given that:

$\sqrt{\sqrt{3-\sqrt{8}}+\sqrt{6+\sqrt{32}}} = \sqrt{ \sqrt{a+b\sqrt{c}}}$

Squaring both side:

$\sqrt{3-\sqrt{8}}+\sqrt{6+\sqrt{32}} = \sqrt{a+b\sqrt{c}}$

Squaring the left hand side:

$(LHS)^2 = 3-\sqrt{8}+2(\sqrt{3-\sqrt{8}})( \sqrt{6+\sqrt{32}}) +6+\sqrt{32}$

$\quad \quad \quad = 9-\sqrt{8}+\sqrt{32} + 2(\sqrt{3-\sqrt{8}}) ( \sqrt{2(3+\sqrt{8})})$

$\quad \quad \quad = 9-2\sqrt{2}+4\sqrt{2} + 2 \sqrt{2} (\sqrt{3-\sqrt{8}}) ( \sqrt{3+\sqrt{8}})$

$\quad \quad \quad = 9 + 2\sqrt{2} + 2 \sqrt{2} \sqrt{9-8}$

$\quad \quad \quad = 9 + 2\sqrt{2} + 2 \sqrt{2} \sqrt{1}$

$\quad \quad \quad = 9 + 4\sqrt{2}$

$\Rightarrow a = 9, \quad b = 4, \quad c = 2$

$\Rightarrow a+b+c = \boxed{15}$

How did you get +2(root 3 - root 8)(root 6 + root 32) in the first line? I see if you square the left hand side, u remove the root on 3, the root on 6. And your left with root 8 and root 32.

Khan Asif - 6 years, 9 months ago

Khan Asif, let $a=\sqrt{3-\sqrt{8}}$ and $b=\sqrt{6+\sqrt{32}}$ , then

$(a+b)^2=a^2+2ab+b^2$

$\quad \quad =3-\sqrt{8}+2(\sqrt{3-\sqrt{8}}) ( \sqrt{6+\sqrt{32}}) +6+\sqrt{32}$

Chew-Seong Cheong - 6 years, 9 months ago

Khan Asif, $(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2 = a^2 + 2ab + b^2 \ne a^2+b^2$

Note that: $5^2 = (3+2)^2 = 3^2+2(3)(2)+2^2 = 25 \ne 3^2+2^2 = 13$

Also: $5^2 = (4+1)^2 = 4^2+2(4)(1)+1^2 = 25 \ne 4^2+1^2 = 17$

Chew-Seong Cheong - 6 years, 9 months ago

Adam Zaim
Sep 20, 2014

$\sqrt { \sqrt { 3-\sqrt { 8 } } +\sqrt { 6+\sqrt { 32 } } } =\sqrt { \sqrt { a+b\sqrt { c } } } \\ \sqrt { 3-\sqrt { 8 } } +\sqrt { 6+\sqrt { 32 } } =\sqrt { a+b\sqrt { c } } \\ { \left( \sqrt { 3-\sqrt { 8 } } +\sqrt { 6+\sqrt { 32 } } \right) }^{ 2 }={ \left( \sqrt { a+b\sqrt { c } } \right) }^{ 2 }\\ 3-\sqrt { 8 } +2\left( \sqrt { 3-\sqrt { 8 } } \right) \left( \sqrt { 6+\sqrt { 32 } } \right) +6+\sqrt { 32 } =a+b\sqrt { c } \\ 9-\sqrt { 8 } +\sqrt { 32 } +2\left( \sqrt { 3-\sqrt { 8 } } \right) \left( \sqrt { 6+\sqrt { 32 } } \right) =a+b\sqrt { c }$

Simplify $2\left( \sqrt { 3-\sqrt { 8 } } \right) \left( \sqrt { 6+\sqrt { 32 } } \right)$

$2\left( \sqrt { 3-\sqrt { 8 } } \right) \left( \sqrt { 6+\sqrt { 32 } } \right) \\ =2{ \left( 3-\sqrt { 8 } \right) }^{ \frac { 1 }{ 2 } }{ \left( 6+\sqrt { 32 } \right) }^{ \frac { 1 }{ 2 } }\\ =2{ \left( \left( 3-\sqrt { 8 } \right) { \left( 6+\sqrt { 32 } \right) } \right) }^{ \frac { 1 }{ 2 } }\\ =2{ \left( 18+3\sqrt { 32 } -6\sqrt { 8 } -\sqrt { 8\ast 32 } \right) }^{ \frac { 1 }{ 2 } }\\ =2{ \left( 18+3\sqrt { 8\ast 4 } -6\sqrt { 8 } -\sqrt { 256 } \right) }^{ \frac { 1 }{ 2 } }\\ =2{ \left( 18+6\sqrt { 8 } -6\sqrt { 8 } -16 \right) }^{ \frac { 1 }{ 2 } }\\ =2\sqrt { 2 }$

So...

$9-\sqrt { 8 } +\sqrt { 32 } +2\left( \sqrt { 3-\sqrt { 8 } } \right) \left( \sqrt { 6+\sqrt { 32 } } \right) =a+b\sqrt { c } \\ 9-\sqrt { 8 } +\sqrt { 32 } +2\sqrt { 2 } =a+b\sqrt { c } \\ 9-\sqrt { 4\ast 2 } +2\sqrt { 2 } +\sqrt { 16\ast 2 } =a+b\sqrt { c } \\ 9-2\sqrt { 2 } +2\sqrt { 2 } +4\sqrt { 2 } =a+b\sqrt { c } \\ 9+4\sqrt { 2 } =a+b\sqrt { c } \\ \therefore a+b+c=9+4+2=15$

Christian Daang
Oct 19, 2014

Simplify First the equation inside the radical,

√(3-√8) + √(6+√32)

= [√(3-√8) + √(6+√32)]^2

= 3-2√2 + 2√2 + 6 + 4√2

= 9 + 4√2

= √(9+4√2)

then, the whole expression is equal to:

√[√(9+4√2)]

where in, a = 9, b=4, c=2

giving us, a+b+c = 9+4+2 = 15

Will you square ???

The answer is 15.

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