(a) A natural number,
(b) A rational number but not a natural number,
(c) An irrational number not exceeding 6,
(d) An irrational number exceeding 6.
Please help with this, i can't manage to simplify it. Thanks in advance.
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@user35508 Why irrational? – dxiv Mar 08 '17 at 06:24
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2Because $x^2$ is not the square of a rational number. – Robert Israel Mar 08 '17 at 06:25
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1Use the double radicals formula to simplify and the it will be apparent that you have an irrational exceeding 6. – Maczinga Mar 08 '17 at 06:26
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Silly question, but how do we know that square root of $226$ will not be a rational number? – Idkwoman Mar 08 '17 at 06:27
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3@RobertIsrael You are right, my arithmetic was off. To the OP: let $a=\sqrt{13+3\sqrt{\frac{23}{3}}}$ and $b= \sqrt{13-3\sqrt{\frac{23}{3}}},$. Then $a^2+b^2=26$ and $ab=10$, so $a+b = \sqrt{a^2+b^2+2ab}=\sqrt{46}$. – dxiv Mar 08 '17 at 06:28
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1For your comment on 226, it lies between the two perfect squares 225 (15 squared) and 256 (16 squared) and so is not a perfect square itself and hence $\sqrt{226}$ isn't rational. – Triatticus Mar 08 '17 at 06:32
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Got it! Many thanks.. – Idkwoman Mar 08 '17 at 06:35
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Sorry Idkwoman but please change your profile picture its very disturbing. – jonsno Mar 08 '17 at 06:36
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4Sorry for your trouble, but that's how my face looks when i'm able to solve a math problem. – Idkwoman Mar 08 '17 at 06:37
1 Answers
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Let us assume $$\sqrt{13+3\sqrt{\frac{23}{3}}} +\sqrt{13-3\sqrt{\frac{23}{3}}}=x $$ Squaring and simplifying gives $$26+20=x^2$$ which gives $$x=+\sqrt{46}$$
Since $6= \sqrt{36}<\sqrt{46}<\sqrt{49}=7$
The result is $(D)$
LM2357
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