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

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- Feb 25th 2011, 03:34 PMstymiedtwo 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 - Feb 25th 2011, 03:41 PMAckbeet
I think you can solve this problem using algebra. What ideas have you had so far?

- Feb 25th 2011, 03:51 PMlanierms
Let's say the two numbers are : $\displaystyle x, y$.

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

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

Factorize :

$\displaystyle t^2-27t+72=0$

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

$\displaystyle t=24$ or $\displaystyle t=3$

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

So the two numbers are $\displaystyle 24, 3$. - Feb 25th 2011, 03:55 PMAckbeet
Looks good to me!

[EDIT] I was confused, thinking that stymied produced post # 3. lanierms' solution is correct. - Feb 25th 2011, 04:28 PMstymied
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! - Feb 25th 2011, 04:42 PMAckbeet
1. The general quadratic is usually written $\displaystyle ax^{2}+bx+c=0,$ the solution of which is

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

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

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

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

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

If you foil out the RHS, you get

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

Therefore, you want $\displaystyle b+a=27,$ and $\displaystyle ab=72.$ Does that make sense? - Feb 25th 2011, 04:56 PMstymied
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 - Feb 25th 2011, 05:01 PMAckbeet
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?

- Feb 25th 2011, 05:12 PMstymied
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 - Feb 25th 2011, 05:19 PMAckbeet
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 $\displaystyle 1\times 72, 2\times 36, 3\times 24, 4\times 18, 6\times 12, 8\times 9.$ For which of those products do the multiplicands sum to 27?

- Feb 25th 2011, 05:22 PMstymied
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 - Feb 25th 2011, 05:26 PMAckbeet
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.