Simple Indicie problem

**Jordan** · Feb 3rd 2008, 04:03 PM

Sorry, very simple question i'm stuck on.

Example in my book shows:

$5^\frac{1}{2}-5^\frac{3}{2}+5^\frac{5}{2} = \sqrt{5} - 5\sqrt{5} + 5^2\sqrt{5} = (1 - 5 + 25)\sqrt{5} = 21\sqrt{5}$

My Q.

$7^\frac{1}{2}-7^\frac{3}{2}+7^\frac{5}{2}$ So I just substituted to get $\sqrt{7} - 7\sqrt{7} + 7^2\sqrt{7} = (1 - 7 + 49)\sqrt{7} = 43\sqrt{7}$

Which is way off the answer in my book of $4\sqrt[3]{7}$

thanks!

**Jhevon** · Feb 3rd 2008, 04:26 PM

Originally Posted by Jordan

Sorry, very simple question i'm stuck on.

Example in my book shows:

$5^\frac{1}{2}-5^\frac{3}{2}+5^\frac{5}{2} = \sqrt{5} - 5\sqrt{5} + 5^2\sqrt{5} = (1 - 5 + 25)\sqrt{5} = 21\sqrt{5}$

My Q.

$7^\frac{1}{2}-7^\frac{3}{2}+7^\frac{5}{2}$ So I just substituted to get $\sqrt{7} - 7\sqrt{7} + 7^2\sqrt{7} = (1 - 7 + 49)\sqrt{7} = 43\sqrt{7}$

Which is way off the answer in my book of $4\sqrt[3]{7}$

thanks!

$43 \sqrt{7}$ is the right answer (you could have checked this with a calculator). the book made a typo. evidently the small 3 should be a big 3

**Soroban** · Feb 4th 2008, 06:28 PM

Hello, Jordan!

$7^\frac{1}{2}-7^\frac{3}{2}+7^\frac{5}{2}$

So I just substituted to get: . $\sqrt{7} - 7\sqrt{7} + 7^2\sqrt{7} \;= \;(1 - 7 + 49)\sqrt{7} \;=\; 43\sqrt{7}$ . . . . Right!

Which is way off the answer in my book: . $4\sqrt[3]{7}$

I bet it's a typo in your book.
. . There is no way that a cube root could appear in that problem!

Look at your answer and theirs: . $43\sqrt{7}\qquad4\sqrt[3]{7}$

It looks like "The Case of the Misplaced 3" (coming to a theater near you).

**Jordan** · Feb 5th 2008, 11:50 AM

Haha, I didn't check it on the calculator because I just assumed the book was right. My mistake.

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