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$\sqrt{57+40\sqrt2} - \sqrt{57-40\sqrt2}$

I'm completely clueless about how to use the formula $a^2 + 2ab + b^2 = (a+b)^2$ to factor the expressions.

With the help of commenters, I successfully factored the expressions.

$\sqrt{57+40\sqrt{2}} = 5+4\sqrt{2}$ and $\sqrt{57-40\sqrt{2}} = 5-4\sqrt{2}$. $5+4\sqrt{2}-4\sqrt{2}+5=10$.

N. F. Taussig
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karina__s
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3 Answers3

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$\sqrt{57+40\sqrt{2}}=\sqrt{25+40\sqrt{2}+32}=\sqrt{5^2+2\cdot 5\cdot 4\sqrt{2}+(4\sqrt{2})^2}=\sqrt{(5+4\sqrt{2})^2}=5+4\sqrt{2}$

eMathHelp
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$$\sqrt{57+40\sqrt2} - \sqrt{57-40\sqrt2} $$ $$\Rightarrow \sqrt{25+32+2.(5).(4\sqrt2}) - \sqrt{25+32-2.(5).(4\sqrt2)}$$ $$\Rightarrow 5+4\sqrt2 - (4\sqrt2-5)$$ $$\Rightarrow 10$$

Swastik Saha
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    Your answer is incorrect. $5 - 4\sqrt{2} = \sqrt{25} - \sqrt{32} < 0$. Therefore, it cannot be the square root of $\sqrt{57 - 40\sqrt{2}}$. – N. F. Taussig Sep 11 '22 at 12:53
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In general, when asked to simplify an expression of the form $$z:=\sqrt{x+a\surd y}\pm\sqrt{x-a\surd y},$$ the trick is to square it, and see what happens. Thus $z^2=\left(\sqrt{x+a\surd y}\pm\sqrt{x-a\surd y}\right)^2$, and so $$z^2=2x\pm2\sqrt{x^2-a^2y}.$$ As usual in this type of problem, $x^2-a^2y$ is a perfect square: in this case, $57^2-40^2\cdot2=49=7^2$. After taking the square root, we end up with $$z=\sqrt{2\cdot57-2\cdot7}=10.$$

John Bentin
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