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Algebra Level 3

$\sqrt{23+\sqrt{140}+\sqrt{308}+\sqrt{220}}$ can be written as $\sqrt{a}+\sqrt{b}+\sqrt{c}$ . What is the value of $a+b+c$ ?

1 solution

Chew-Seong Cheong
Feb 7, 2015

$\sqrt{23+\sqrt{140}+\sqrt{308}+\sqrt{220}} = \sqrt{a}+\sqrt{b}+\sqrt{c} = \sqrt {( \sqrt{a}+\sqrt{b}+\sqrt{c})^2}$

$=\sqrt{a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})} = \sqrt{\boxed{23}+2(\sqrt{35}+\sqrt{77}+\sqrt{55}})$

$= \sqrt{5+7+11+2(\sqrt{5\dot{}7}+\sqrt{7\dot{}11}+\sqrt{11\dot{}5}})$

$\Rightarrow a+b+c = \boxed{23}$

but how 2[{(35)^1/2)}+{(77)^1/2}+{(55)^1/2)}]= [{(ab)^1/2)}+{(bc)^1/2}+{(ca)^1/2)}]

Gaurav Purswani - 6 years, 4 months ago

$\sqrt{140}+\sqrt{308}+\sqrt{220} = 2\sqrt{35}+2\sqrt{77}+2\sqrt{55}$

$= 2(\sqrt{5\dot{}7}+\sqrt{7\dot{}11}+\sqrt{11\dot{}5})$

$\Rightarrow a,b.c = 5,7,11$

Chew-Seong Cheong - 6 years, 4 months ago