I need help with this problem:

ab=2.4(a+b)

a=√(6^2-d^2)

b=√(9^2-d^2)

redeem d from:

(√(36-d^2))(√(81-d^2))=2.4(√(36-d^2)+√(81-d^2))

d=?????

(Sorry if my english is not so good. I'm from Sweden)

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- Oct 25th 2007, 05:31 AM #1

- Joined
- Oct 2007
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## [SOLVED] Need help now!

I need help with this problem:

ab=2.4(a+b)

a=√(6^2-d^2)

b=√(9^2-d^2)

redeem d from:

(√(36-d^2))(√(81-d^2))=2.4(√(36-d^2)+√(81-d^2))

d=?????

(Sorry if my english is not so good. I'm from Sweden)

- Oct 25th 2007, 06:39 AM #2

- Oct 25th 2007, 08:18 AM #3

- Joined
- Oct 2007
- Posts
- 3

- Oct 26th 2007, 07:54 AM #4
ok. first, you need to square both sides of the equation and apply the laws of exponents..

then, isolate the one with the radical on one side of the equation, then square it again..

you must notice that your equation would be of degree 4, but it is just like a quadratic form by letting x=d^2.

then solve for x. after solving for the values of x, equate it to d^2 and solve for d..

you might get 4 solutions for d but others are just extraneous..