word problems

**Chris85** · September 17th 2007, 08:09 AM

Hi all, i'm after a bit of advice, problem is i'm not that great with word problems i am stuck on a couple and i am wodering if anyone has any tips for solving word problems and forming equations from them?? Here is one i'm a bit stuck on.

The cost of an anorak rose by £6. As a result a shop could buy 5 fewer anoraks for £600. If the cost of the anorak was £x before the rise, find expressions, in terms of x,for the number of anoraks which could be bought before and after the rise. Hence form an equation in x and show that it reduces to x^2+ 6x-720=0. solve this equation and state the original cost of the anorak.

can somebody tell me how i should go about doing these??

Thanks,
Chris.

**DivideBy0** · September 17th 2007, 08:27 AM

We know x is the original cost per anorak.
Let a be the amount of anoraks purchased.
Let T = total cost

From this we have

$ax=T$

Now, originally, you would have:

$ax=600$

But now, cost has risen by six, so amount must decrease by five. Remember that the new cost * new amount still = 600.

$(a-5)(x+6)=600$

But what is a? Well, we can put a in terms of x.

$ax=600$

$a=\frac{600}{x}$

So we have the equation:
$\left(\frac{600}{x}-5 \right)(x+6)=600$

Hopefully you can take it from there

**Chris85** · September 17th 2007, 12:14 PM

Hello thanks, i see what your done. How would i go about solving the final equation at the bottom for the price of the anorak? I played around and got £24 i think thats right but it was trial and error. So as i say how would i solve the equation at the bottom properly?

Thankyou for your help

Chris.

**DivideBy0** · September 17th 2007, 06:34 PM

$\left( \frac{600}{x}-5 \right)(x+6)=600$

$600+\frac{3600}{x}-5x-30=600$

$-5x-30+\frac{3600}{x}=0$

$-5x^2-30x+3600=0$

$x^2+6x-720=0$

You have to find two numbers with multiply to give -720 and add to give 6. Since -720 is negative, you will need a negative and a positive number. What you could do is list the factors of 720:
(1,720), (2,360), (3,240), (4,180), (5,144), (6,120), (8,90), (9,80), (10,72), (12,60), (15,48), (16,45), (18,40), (20,36), (24,30)
And test them to see if they fit the requirements. Here, 24 and 30 do.

$(x-24)(x+30)=0$

$x-24=0$ or $x+30=0$ (Null Factor Law)

$x=24$ or $x=-30$

Of course this is a very long and hard way to do the problem (especially for 720, as it has loads of factors), but there are easier ways.

Another way is 'completing the square':

$x^2+6x-720=0$

$x^2+6x+(\frac{6}{2})^2-720=(\frac{6}{2})^2$ (add the square of half the coefficient of the x-term to both sides)

$x^2+6x+9=729$ (notice the left hand side is a perfect square)

$(x+3)^2=729$ (factorise the LHS)

$x+3=\pm 27$

$x=-3 \pm 27 = -30,24$

But perhaps the easiest way for this problem, is to plug the values into the quadratic formula. For any equation $ax^2+bx+c=0$ , we have

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In this case we have a = 1, b = 6, c = -720

$x=\frac{-(6) \pm \sqrt{(6)^2-4(1)(-720)}}{2(1)}$

$x=\frac{-6 \pm \sqrt{2916}}{2}$

$x=\frac{-6 \pm 54}{2}$

$x=-3 \pm 27=24,-30$

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