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survivorboi · Jul 31, 2009, 10:31 AM

The sum of the squares of two positive integers is 193. The product
of the two integers is 84. What is the sum of the two integers?

How do I solve it?

Unknown008 · Jul 31, 2009, 10:36 AM

Assign letters to the unknown integers.

Let x be the first one, and y the second one.

From the first piece of info;

$x^2+y^2=193$

From the second piece of info;

$xy=84$

It's a simultaneous equation. Find their values, then you'll be able to write down the sum of the two integers. :)

survivorboi · Jul 31, 2009, 11:21 AM

But the problem is that I haven't studied quadratic equations yet. How then, I know what is xy=84?

84/x=y?

Unknown008 · Jul 31, 2009, 12:16 PM

Oh, I thought you understood these... sorry.. :o

Ok, here we go! :)

$x^2 + y^2 = 193$

$xy=84$

Yes, start by making one of them the subject of formula.

$y = \frac{84}{x}$

Then, replace it in the first equation. You now know the 'equivalent' of y, replace all y by that equivalent:

$x^2 + (\frac{84}{x})^2 = 193$

$x^2 + \frac{7056}{x^2} = 193$

Multiply everything by x^2 to remove fractions;

$x^4 + 7056 = 193x^2$

Now rearrange to the same side:

$x^4 - 193x^2+ 7056 = 0$

$(x^2-49)(x^2-144)=0$

Therefore, x = 7 or 12.

And so is y.

And the sum of the integers is (12 + 7)= 19!

:)

survivorboi · Jul 31, 2009, 06:12 PM

$x^4 - 193x^2+ 7056 = 0$

$(x^2-49)(x^2-144)=0$

Therefore, x = 7 or 12.

And so is y.

And the sum of the integers is (12 + 7)= 19!

Umm, please go back and explain how you got from x^4 to x^2 and how you got 49 and 144 please. Where did 7056 goo?

galactus · Aug 1, 2009, 11:21 AM

Rewrite it as $x^{2}-193x+7056$

Then, factor.

What two numbers when multiplied equal 7056 and when added equal -193?

How about -144 and -49

(-144)(-49)=7056

-144+(-49)=-193

$x^{2}-144x-49x+7056$

$(x^{2}-144x)-(49x-7056)$

$x(x-144)-49(x-144)$

$(x-49)(x-144)$

Now, because it was an x^4 in the beginning, put an x^2 back in place of the x

$(x^{2}-49)(x^{2}-144)$

Note these are the difference of two squares:

$(x^{2}-7^{2})(x^{2}-12^{2})$

But $x^{2}-y^{2}=(x+y)(x-y)$

So, we have $(x-7)(x+7)(x-12)(x+12)$

Unknown008 · Aug 1, 2009, 11:32 AM

To add to galactus explanation, I'll put it like this.

You let for example $a=x^2$

What will you have?

$x^4 - 193x^2 + 7056 = 0$

If a = x^2, then a^2 = x^4. So;

$a^2 - 193a + 7056 = 0$

Then factorise normally, to have:

$(a-49)(a-144)=0$

Then, replace back to x^2, you have:

$(x^2-49)(x^2-144)=0$

And, solving, you have:

$x^2-49=0$

$x^2=49$

$x=\pm \sqrt{49} = \pm7$

But since the integers are only positive, you consider only 7.

The same thing goes for the other integer. :)

survivorboi · Aug 7, 2009, 11:15 AM

THanks guys =)

Unknown008 · Aug 8, 2009, 12:28 AM

You're welcomed suvivorboi! Care to take a look at my thread? There are lots of challenging questions! :)

https://www.askmehelpdesk.com/mathem...ns-384108.html