two numbers

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February 25th 2011, 03:34 PM
stymied

two numbers

Hi there
wondering if this can be solved using Algebra?
the product of two numbers is 72 and their sum is 27.
I am interested in the steps to arrive at the numbers 24 and 3.
cheers
February 25th 2011, 03:41 PM
Ackbeet

I think you can solve this problem using algebra. What ideas have you had so far?
February 25th 2011, 03:51 PM
lanierms

Let's say the two numbers are : $x, y$ .

Then, $xy=72$ , $x+y=27$

$x, y$ are two roots of a quadratic equation : $t^2-27t+72=0$

Factorize :
$t^2-27t+72=0$

$(t-24)(t-3)=0$

$t=24$ or $t=3$

$x=24, y=3$ or $x=3, y=24$

So the two numbers are $24, 3$ .
February 25th 2011, 03:55 PM
Ackbeet

Looks good to me!

[EDIT] I was confused, thinking that stymied produced post # 3. lanierms' solution is correct.
February 25th 2011, 04:28 PM
stymied

Thanks for the quick replies which have thrown up a couple of questions for me.
1. In a quadratic equation is it always ax squared minus bx + c =0 or can the minus be a plus? does it matter?
2. I can't seem to work out how in the factorizing you got from the 1st line to the second
thanks again for your help!
February 25th 2011, 04:42 PM
Ackbeet

1. The general quadratic is usually written $ax^{2}+bx+c=0,$ the solution of which is

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

If you wrote your quadratic as $ax^{2}-bx+c=0,$ then you'd merely flip the sign of $b$ in the solution:

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

2. When you're factoring a monic quadratic like this one (that is, the coefficient of $t^{2}$ is 1), your goal is to find two numbers a, b, such that

$t^{2}-27t+72=(t-a)(t-b).$

If you foil out the RHS, you get

$t^{2}-bt-at+ab=t^{2}-(b+a)t+ab.$

Therefore, you want $b+a=27,$ and $ab=72.$ Does that make sense?
February 25th 2011, 04:56 PM
stymied

I am stiil not sure how you go from t squared minus 27 t plus 72 equals zero.
to putting in the numbers 24 and 3? Do you just have to work out these numbers in your head or is there a way to foil them out from the first line?
(Sorry I am a bit slow on the math front!)
cheers
February 25th 2011, 05:01 PM
Ackbeet

Well, factoring is a bit like a puzzle. Since everything in sight's an integer, I'd look at the factors of 72. What are the factors of 72?
February 25th 2011, 05:12 PM
stymied

OK I see what your saying I thought there may be a process that you could use to drill down to the point where you could get a difinitive x = 24, y =3 or vise versa.
I still can't see how lanierms above went from the first to second line putting in 24 and 3, did he just think of the factors of 72 and work it out in his head?
This is great having this conversation and I appreciate your time!
cheers
February 25th 2011, 05:19 PM
Ackbeet

Quote:

Originally Posted by stymied

OK I see what your saying I thought there may be a process that you could use to drill down to the point where you could get a difinitive x = 24, y =3 or vise versa.
I still can't see how lanierms above went from the first to second line putting in 24 and 3, did he just think of the factors of 72 and work it out in his head?

Probably. The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. You need to take two of those, multiplied together, to get 72. So your options are $1\times 72, 2\times 36, 3\times 24, 4\times 18, 6\times 12, 8\times 9.$ For which of those products do the multiplicands sum to 27?
February 25th 2011, 05:22 PM
stymied

Hi AcKbeet I just worked it out! Using your formula above for finding x from a quadratic.
Thanks heaps for your time. One of my sons has just started high school and we try and work things out together so I am sure I''ll be in touch in the future!
This has been a great forum for me. Its just after lunch on a Saturday in Australia.
All the best
February 25th 2011, 05:26 PM
Ackbeet

Yeah, for a quadratic, I tend to prefer using the quadratic formula. There are times when it's better to factor, but solving quadratics is a fairly routine matter, in the grand scheme of things, so an algorithmic process I don't have to think about too much appeals to me.

You're very welcome. It's almost time to get some shut-eye here in Connecticut, USA.