Quadratic help

**sinjid9** · April 22nd 2010, 01:14 PM

A store owner buys a quantity of balls for $600. If they had each cost $0.25 less, she would have 10 more for the same money. How much did she pay for each ball.

(x-0.25)*(y+10)
is how far I got. Is this even a quadratic equation?

**Archie Meade** · April 22nd 2010, 02:15 PM

Originally Posted by sinjid9

A store owner buys a quantity of balls for $600. If they had each cost $0.25 less, she would have 10 more for the same money. How much did she pay for each ball.

(x-0.25)*(y+10)
is how far I got. Is this even a quadratic equation?

Hi sinjid9,

the amount of money is the same in both cases and is (price)(quantity).

Hence, if x=price, y=number of balls

$xy=(x-0.25)(y+10)=600$

Now write one variable in terms of the other using $xy=600$

$x=\frac{600}{y},\ or\ y=\frac{600}{x}$

This leads to the quadratic in a single variable.

For example....

$(x-0.25)\left(\frac{600}{x}+10\right)=600$

$\frac{1}{x}(x-0.25)(600+10x)=600$

$(x-0.25)(600+10x)=600x$

$600x+10x^2-150-2.5x=600x$

$10x^2-2.5x-150=0$

$100x^2-25x-1500=0$

$4x^2-x-60=0$

$(4x+15)(x-4)=0$

$x-4=0$ as x has to be positive.

$x=4\ dollars,\ y=150$

notice $(4-0.25)(150+10)=(3.75)(160)=600$

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