someone help!

**ify00** · May 17th 2006, 06:10 AM

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.

**earboth** · May 17th 2006, 11:18 AM

Originally Posted by ify00

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.

Hello,

keep in mind:

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

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

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

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

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

Greetings

EB

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someone help!