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by **EdSewersPark** on Mon May 23, 2011 9:13 pm

$(2^{1/2}+2^{3/2})^2$

This one looks so easy, but still I am getting it wrong. Solution is 18 and I keep getting 10.

**Sponsor**

by **stapel_eliz** on Mon May 23, 2011 10:44 pm

EdSewersPark wrote:Solution is 18 and I keep getting 10.

It's a lot easier to check work that we can see... :wink:

by **EdSewersPark** on Tue May 24, 2011 12:36 am

$(2^{1/2}+2^{3/2})^2$

$(2^{2/2}+2^{6/2})$ or $(2+2^{6/2})$ or $(2+\sqrt2^6)$

$(2+\sqrt64)$

$(2+8)=10$

by **stapel_eliz** on Tue May 24, 2011 10:48 am

EdSewersPark wrote: $(2^{1/2}+2^{3/2})^2$

$(2^{2/2}+2^{6/2})$ ...

Hint: (a + b)² does not equal a² + b²! :wink:

Instead, try using properties of radicals:

. . . . . $2^{\frac{1}{2}}\, +\, 2^{\frac{3}{2}}\, =\, \sqrt{2}\, +\, \sqrt{2^3}\, =\, \sqrt{2}\, +\, 2\sqrt{2}\, =\, \sqrt{2}\left(1\, +\, 2\right)$

...and so forth. :wink:

by **EdSewersPark** on Tue May 24, 2011 12:13 pm

$(\2^{1/2}+2^{3/2})$

$(\sqrt2+\sqrt2^3)(\sqrt2+\sqrt2^3)$

$(\sqrt2\sqrt2+\sqrt2\sqrt8+\sqrt2\sqrt8+\sqrt8\sqrt8)$

$(2+\sqrt16+\sqrt16+8)$

$(2+4+4+8) = 18$

tada. Thank you for your help!

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simplifying [2^{1/2} + 2^{3/2}]^2

simplifying [2^{1/2} + 2^{3/2}]^2

Re: simplifying [2^{1/2} + 2^{3/2}]^2

Re: simplifying [2^{1/2} + 2^{3/2}]^2