3
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We have an expression such as

 Sqrt[-2 - d^2 + 2 Sqrt[1 + d^2]] Sqrt[-d^2 - 2 (1 + Sqrt[1 + d^2])]

enter image description here

When we use

FullSimplify[Sqrt[-2 - d^2 + 2 Sqrt[1 + d^2]] Sqrt[-d^2 - 2 (1 + Sqrt[1 + d^2])] ]

The result is nothing simplified.

But when we use

FullSimplify[(-2 - d^2 + 2 Sqrt[1 + d^2]) (-d^2 - 2 - 2 Sqrt[1 + d^2])]

in which we firstly expanded the second expression and after that we multiplied them to each other albeit without the Sqrt.

we can see the result is d^4 and Sqrt[d^4] is d^2.

How can we directly obtain the the most simplified yield from the first expression i.e.,

Sqrt[-2 - d^2 + 2 Sqrt[1 + d^2]] Sqrt[-d^2 - 2 (1 + Sqrt[1 + d^2])]
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2 Answers 2

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Your purported solution isn't correct. For example:

Sqrt[-2-d^2+2 Sqrt[1+d^2]] Sqrt[-d^2-2 (1+Sqrt[1+d^2])] /. d->.1

-0.01 + 0. I

It helps to add assumptions:

FullSimplify[
    Sqrt[-2-d^2+2 Sqrt[1+d^2]] Sqrt[-d^2-2 (1+Sqrt[1+d^2])],
    d ∈ Reals
]

-d^2

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  • $\begingroup$ @ Carl Woll, thaks, but when I use Sqrt[-2-d^2+2 Sqrt[1+d^2]] Sqrt[-d^2-2 (1+Sqrt[1+d^2])] /. d->.1 I just see 0.01!!!!!!!! and not what you have seen!! $\endgroup$ May 14, 2017 at 12:11
  • $\begingroup$ And Sqrt[-2 - d^2 + 2 Sqrt[1 + d^2]] Sqrt[-d^2 - 2 (1 + Sqrt[1 + d^2])] /. d -> 1/10 // Simplify evaluates to -1/100 $\endgroup$
    – Bob Hanlon
    May 14, 2017 at 14:59
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In:

expr = Sqrt[-2 - d^2 + 2 Sqrt[1 + d^2]] Sqrt[-d^2 - 2 (1 + Sqrt[1 + d^2])];
Assuming[d ∈ Reals, Simplify[expr]]

Out:

-d^2
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2
  • $\begingroup$ Simplify[expr, d \[Element] Reals] suffices. $\endgroup$
    – corey979
    May 14, 2017 at 8:29
  • $\begingroup$ @corey979 I don't know why it doesn't work sometimes, that's why I use Assuming instead. $\endgroup$
    – webcpu
    May 14, 2017 at 8:53

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