What is the value of $$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$$
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Yeah it is correct as written and the answer is actually 4 – siva_sai_kumar_reddy Jul 19 '17 at 14:01
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The root at the first is for the entire first value – siva_sai_kumar_reddy Jul 19 '17 at 14:02
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Could you please provide some details about your own attempts at solving the problem? – ekkilop Jul 19 '17 at 14:11
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Using a calculator, $4$. (You wanted the value, right ?) – Jul 19 '17 at 14:20
2 Answers
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With $x=\sqrt{7+2\sqrt{12}}$ and $y=\sqrt{7-2\sqrt{12}}$, we have $x^2+y^2=14$ and $xy=\sqrt{7^2-2^2\cdot12}=1$, so $(x+y)^2=14+2\cdot1=16$ and thus $x+y=4,$ since $x,y$ are positive.
In general, we'd have $$\sqrt{a+b\sqrt{c}}+\sqrt{a-b\sqrt{c}}=\sqrt{2a+2\sqrt{a^2-b^2c}}.$$
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Since $(2\pm\sqrt3)^2=7\pm4\sqrt3=7\pm2\sqrt{12}$,$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}=2+\sqrt3+2-\sqrt3=4.$$
José Carlos Santos
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@user162748 I was looking for $a,b\in\mathbb N$ such that $(a+b\sqrt3)^2=7+4\sqrt3$. But $(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt3$. So, I wanted to have $a^2+3b^2=7$ and $ab=2$ and then I took $a=2$ and $b=1$. – José Carlos Santos Jul 19 '17 at 14:12
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@user162748 You can also use this very smple radical denesting rule. – Bill Dubuque Jul 19 '17 at 16:42