View Full Version : Quadratic Equation involving money
HuiLi
Feb 12, 2010, 06:52 PM
A trader bought some paraffin for $500.He paid $x for each litre of paraffin.
Qn:
a)Due to a leak,he lost 3 litres of paraffin.He sold the remainder of the paraffin for $1 per litre more than he paid for it.Write down an expression,in terms of x,for the sum of money he received.
b)He made a profit of $20.Write an equation in x to represent this information.
galactus
Feb 13, 2010, 11:37 AM
Since he paid $500 for all the parafin and paid $x for each liter, then he bought 500/x liters.
But 3 leaked out and he sold it for $1 more than what he paid.
(x+1)(\frac{500}{x}-3)
Now, you try to do part b. If he made $20 profit, then how much did he gross? Set the expression equal to that and solve for x.
HuiLi
Feb 14, 2010, 12:27 AM
I think I get something wrong here...
The equation I get:
523xsquare+23x-500=0
The answer:
3xsquare+23x-500=0
I can't find the problem with it
galactus
Feb 14, 2010, 04:57 AM
You made a mistake then. What I showed you reduces to 3x^{2}+23x-500=0
Just expand
(x+1)(\frac{500}{x}-3)-520=0
and multiply by -x.
Post how you came up with what you have and maybe we can help. As is, we have no way of knowing how you got that.
HuiLi
Feb 14, 2010, 08:52 PM
Here is how I get it:
(x+1)(500/x-3)=520
(500-3x(x+1)/xsquare+x)-520=0
500x+500-3xsquare-3x/xsquare+x
(497x+500-3xsqiare/xsquare+x)-520=0
(497x+500-3xsqiare/xsquare+x)-(520xsquare+520x)/xsquare+x
both sides times -x
23x-500+523xsquare/-xsquare-x =0
How do I make it an equation from here on?
HuiLi
Feb 17, 2010, 04:19 AM
(520/x-3)(x-1)
520=(500x+500/xsquare+x)-3x+3
520square+520x=500x+500-3x+3
520xsquare+520x-503x+497=0
520xsquare+17x+497=0
this is the new method I can think of,but I still cannto get the answer?is anyone here?
Unknown008
Feb 17, 2010, 07:28 AM
I think you have got a little problem in expansion:
(x+1)(\frac{500}{x}-3) = 520
I'll take only the left hand side for now.
\begin{array}{rcl}
(x+1)(\frac{500}{x}-3) &=& (x+1)(\frac{500-3x}{x}) \\ &=&
\frac{(x+1)(500-3x)}{x} \\ &=&
\frac{(500x+500-3x^2-3x)}{x} \\ &=&
\frac{(497x+500-3x^2)}{x}\end{array}
Now, I put back the 520 from before:
\begin{array}{rcl}
\frac{(497x+500-3x^2)}{x}&=& 520 \\
497x+500-3x^2 &=& 520x\\
0 &=& 3x^2 + 23x - 500\end{array}
HuiLi
Feb 19, 2010, 05:09 AM
Thanks!that is very clear & helpful!
I finally understand maths & what I did wrong!yay for me :D
Unknown008
Feb 19, 2010, 07:34 AM
You're welcome :)