Simplifying $\sqrt {11 – 4 \sqrt 7} – \sqrt {8 – 2 \sqrt 7}$

algebra-precalculusradicals

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

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.

Best Answer

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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