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?

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. :)

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

84/x=y?

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!

:)

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?

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)

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. :)

THanks guys =)

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

https://www.askmehelpdesk.com/mathematics/amc-questions-384108.html