• Apr 12th 2006, 06:18 AM
dgolverk
I can't understand how I can find the X's here...
(x-4)(30/x+2)=30

Thanks
• Apr 12th 2006, 06:29 AM
CaptainBlack
Quote:

Originally Posted by dgolverk
I can't understand how I can find the X's here...
(x-4)(30/x+2)=30

Thanks

Expand the brackets:

$\displaystyle (x-4)(30/x+2)=30+2x-\frac{120}{x}-8=30$.

Simplify a bit:

$\displaystyle 2x-\frac{120}{x}-8=0$.

Multiply through by $\displaystyle x$:

$\displaystyle 2x^2-8x-120=0$.

Now use the quadratic formula to solve for $\displaystyle x$

RonL
• Apr 12th 2006, 07:06 AM
dgolverk
well, it's the same way that I tried, but my answer isn't right according the answers book - it says that X=10 and X=-6
the book's answer is wrong? or I did something not right?
• Apr 12th 2006, 07:54 AM
earboth
Quote:

Originally Posted by dgolverk
well, it's the same way that I tried, but my answer isn't right according the answers book - it says that X=10 and X=-6
the book's answer is wrong? or I did something not right?

Hello,

Greetings

EB
• Apr 12th 2006, 08:07 AM
earboth
Quote:

Originally Posted by dgolverk
well, it's the same way that I tried, but my answer isn't right according the answers book - it says that X=10 and X=-6
the book's answer is wrong? or I did something not right?

Hello,

that's what you should have done: Use the quadratic formula:

$\displaystyle x=\frac{-(-8) \pm \sqrt{64-4 \cdot 2 \cdot(-120)}}{2\cdot 2}$

$\displaystyle x=\frac{-(-8) \pm \sqrt{64+960}}{2\cdot 2}$

$\displaystyle x=\frac{-(-8) \pm 32}{2\cdot 2}$

$\displaystyle x=\frac{40}{4}\ \vee \ x=\frac{-24}{4}$

Greetings

EB
• Apr 12th 2006, 08:59 AM
dgolverk
intersting...
I did exactly what you have done and got other answers..I guess I forgot something :rolleyes:
Thanks a lot!