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 tony39531 Posts: 1, Reputation: 1 New Member #1 Dec 17, 2007, 08:20 AM
How do I solve a word dilemma using quadratic equation
I have worked for days on this and went to every website and test I could find but cannot wrap my brain around this.

Word problems give me great difficulty and I am supposed to construct a quadratic equation for this. I'd really appreciate some help.

if a retailer spends $48 to purchase some items and two of them were broken in the store and the remaining mugs were sold for$3 above the cost of the mugs, profit was $22 I have to construct an equation to solve for the number of mugs that were originally purchased. I think the formula is supposed to be ax (sq) +bx+c=0 but I have no idea how to plug in the numbers or if this is a proper formula. I have worked for days and even looked at wikipedia and can't get it. Please help :confused:  galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Dec 17, 2007, 05:52 PM Profit = Revenue - Cost. The Profit is$22. The cost is $48. Now, the trickiest part. After 2 are broken, the mugs sell for$3 more, a piece, then what they paid for them.

Each mug cost 48/n dollars. There are n-2 of them top sell.

$\overbrace{22}^{\text{profit}}=\underbrace{(\frac{ 48}{n}+3)(n-2)}_{\text{revenue}}-\overbrace{48}^{\text{cost}}$

The quadratic is $3n^{2}-28n-96=0$

Solve away. One solution will be negative, therefore, erroneous.

Factor, use the quadratic formula, complete the square. Whatever.

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