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Is this solution correct ? please click here to see the question & my answer Sorry for the nasty handwriting (in image). $$\sqrt{a}(2a^2-4/a)\Rightarrow a^{1/2}(2a^2-4a^{-1})\Rightarrow 2(a^{5/2}-2a^{-1/2})\Rightarrow 2\sqrt{a^5-2/a}$$ $$\Rightarrow 2\sqrt{(a^6-2)/a}\Rightarrow 2(a^3/\sqrt{a}-2/\sqrt{a})\Rightarrow 2(6-2/\sqrt{a})$$

Thank you so much! :)

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  • $\begingroup$ I don't particularly follow your next-to-last step. How is $\sqrt{a^6 - 2} = a^3 - 2$? $\endgroup$ – Eevee Trainer Apr 6 at 9:30
  • $\begingroup$ Sorry, it should be $a^3$ - $\sqrt{2}$ $\endgroup$ – Deepak S.M Apr 6 at 9:34
  • $\begingroup$ But what is the question? You have to simplify the provided expression? $\endgroup$ – pluton Apr 6 at 9:35
  • $\begingroup$ @pluton yes, the question is to simplify it. $\endgroup$ – Deepak S.M Apr 6 at 9:36
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Actually, following up on my comment, I realized in general a good bit is wrong. The square root operation does not "distribute" over addition and subtraction, i.e.

$$\sqrt{x+y} \ne \sqrt{x} + \sqrt y$$

Similarly, in general for powers,

$$(x+y)^a \ne x^a + y^a$$

You use this assumption several times in your solution. For example, you claim

$$a^{5/2} - 2a^{-1/2} = \sqrt{a^5 - 2a^{-1}}$$

when in reality you can only say

$$a^{5/2} - 2a^{-1/2} = \sqrt{a^5} -2 \cdot \sqrt{a^{-1}}$$

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  • $\begingroup$ Thanks a lot for your effort! :D $\endgroup$ – Deepak S.M Apr 6 at 9:42
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It is by expanding $$2a^{2+\frac{1}{2}}-4a^{-1}\cdot a^{1/2}$$ and by simplifying $$2a^{5/2}-4a^{-1/2}$$

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