$\sqrt{9+6\sqrt{2}}$ to find length
But how do I express the above in the form of $\sqrt{c} + \sqrt{d}$.
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2$\begingroup$ Try to see what happens computing $(\sqrt{c}+\sqrt{d})^2$; what do they need to be to get what you want? $\endgroup$– pjs36Sep 21, 2015 at 22:57
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$\begingroup$ With what restrictions on $c$ and $d$ ? Without restrictions, you just take $d=0$ and $c= 9+6\sqrt{2}$. $\endgroup$– lhfSep 22, 2015 at 0:06
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$\begingroup$ You can find several similar question on this site. For example, this one. $\endgroup$– Martin SleziakMar 25, 2016 at 9:59
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3 Answers
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Hint: $$9+6\sqrt{2}=6+6\sqrt{2}+3=(\sqrt{6}+\sqrt{3})^2$$
answered Sep 21, 2015 at 22:57
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Let the side of the square be $(\sqrt c+\sqrt d)$ then its area is given as $$(\sqrt c+\sqrt d)^2=9+6\sqrt 2\tag 1$$ $$c+d+2\sqrt{cd}=9+6\sqrt 2$$ by comparing the corresponding rational & irrational parts, we get $$c+d=9\tag 2$$ $$2\sqrt{cd}=6\sqrt 2\tag 3$$ Now, we know $$(\sqrt c-\sqrt d)^2=(\sqrt c+\sqrt d)^2-2\sqrt{cd}$$$$=(9+6\sqrt 2)-2(6\sqrt 2)$$$$(\sqrt c-\sqrt d)^2=9-6\sqrt 2\tag 4$$ multiplying (1) & (4) we get $$(\sqrt c+\sqrt d)^2(\sqrt c-\sqrt d)^2=(9+6\sqrt 2)(9-6\sqrt 2)$$ $$(c-d)^2=9\implies c-d=\pm 3$$ 1. taking positive sign, we get $c-d=3\tag 5$ solving (2) & (5), we get $$\color{red}{c=6, \ d=3}$$ 2. taking negative sign, we get $c-d=-3\tag 6$ solving (2) & (6), we get $$\color{red}{c=3, \ d=6}$$ Both the above cases give the same result. Hence, the side of the square is $$\color{red}{\sqrt c+\sqrt d}=\color{blue}{\sqrt 6+\sqrt 3}$$ or $$\color{red}{\sqrt c+\sqrt d}=\color{blue}{\sqrt 3+\sqrt 6}$$
answered Sep 21, 2015 at 23:21
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$\begingroup$ Why can you compare just the corresponding rational an irrational parts? $\endgroup$– lhfSep 22, 2015 at 0:03
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$\begingroup$ Because it is an equality/equation. So corresponding rational & irrational parts should be equal $\endgroup$ Sep 22, 2015 at 0:07
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2$\begingroup$ From (3) we get $cd=18$ and we can then solve a simple quadratic for $c,d$ by inspection. $\endgroup$– lhfMar 25, 2016 at 10:12
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Notice, let the side of the square be $a$ then its area is given as $$a^2=9+6\sqrt 2$$ $$a^2=9+2(3)\sqrt 2$$ $$a^2=9+2(\sqrt 3\sqrt 3)\sqrt 2$$ $$a^2=9+2(3)\sqrt 2$$ $$a^2=9+2(\sqrt 3\sqrt 2)(\sqrt 3)$$ $$a^2=6+3+2\sqrt{6}\sqrt 3$$ $$a^2=(\sqrt 6)^2+(\sqrt 3)^2+2(\sqrt{6})(\sqrt 3)$$ Using $a^2+b^2+2ab=(a+b)^2$, we get $$a^2=(\sqrt 6+\sqrt 3)^2$$ hence, side of square $$a=\color{red}{\sqrt6+\sqrt 3}$$
answered Sep 21, 2015 at 23:01