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

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

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

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

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

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Solution is 18 and I keep getting 10.
It's a lot easier to check work that we can see...

EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

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

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

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

$(2+\sqrt64)$

$(2+8)=10$

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
$(2^{1/2}+2^{3/2})^2$

$(2^{2/2}+2^{6/2})$...
Hint: (a + b)2 does not equal a2 + b2!

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.

EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

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

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

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

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

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

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

tada. Thank you for your help!

Return to “Intermediate Algebra”