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$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$

I'm having trouble solving this equation. I've tried squaring both sides and got this

$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$

after simplifying

$$19 - 6\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} = a^2$$

and that's where I got stuck.

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    $\begingroup$ Hint: $11-4\sqrt{7}=2^2-2\cdot2\cdot\sqrt{7}+7=(\sqrt{7}-2)^2$. What can you say about $8-2\sqrt{7}$? $\endgroup$ – Stinking Bishop Nov 28 '20 at 16:18
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Hint:

$11=(\sqrt7)^2+2^2$ and $\sqrt7-2>0$

$8=(\sqrt7)^2+1^2$ and $\sqrt7-1>0$

Finally

$$\sqrt{a^2+b^2-2ab}=|a-b|=a-b\text{ if } a-b\ge0$$

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There is a formula to simplify nested radicals like that $$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ It is effective only if $(a^2-b)$ is a perfect square. In your example both of them have this property $$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$ $$\sqrt {11 - \sqrt {112}} - \sqrt {8 -\sqrt {28}} = a$$ $$\sqrt {11 - \sqrt {112}} =\sqrt{\frac{11+\sqrt{121-112}}{2}}-\sqrt{\frac{11-\sqrt{121-112}}{2}}=\sqrt{\frac{11+3}{2}}-\sqrt{\frac{11-3}{2}}=$$ $$=\sqrt 7-2$$ $$\sqrt {8 -\sqrt {28}} =\sqrt{\frac{8+6}{2}}-\sqrt{\frac{8-6}{2}}=\sqrt 7-1$$ so finally $$\sqrt 7-2-(\sqrt 7-1)=-1$$


There is a similar formula for the sum $$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$

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