The equation 2k³ - 4k^(3/2) + 1 = 0

Subsitute p² for k³ to find the value of k.

i got 2p²-4p + 1, and it doesnt factorise to give k=0.44, which is at the back of the book!

i even used the discriminant thing.

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- May 17th 2006, 06:10 AM #1

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- Apr 2006
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- May 17th 2006, 11:18 AM #2Originally Posted by
**ify00**

keep in mind:

$\displaystyle p^2=k^3 \Longrightarrow k=\sqrt[3]{p^2}$

$\displaystyle 2p^2-4p+1=0\Longleftrightarrow p=1+{1\over2}\sqrt{2}\ \vee \ p=1-{1\over2}\sqrt{2}$

Now plug in these values of p into the equation to calculate k and you'll get:

$\displaystyle k=\sqrt[3]{\left( 1+{1\over2}\sqrt{2} \right)^2} \approx 1.4283...$ or

$\displaystyle k=\sqrt[3]{\left( 1-{1\over2}\sqrt{2} \right)^2} \approx .4410...$

Greetings

EB