• February 12th 2010, 01:10 PM
Tessarina
Two integers differ by 12 and the sum of their squares is 74. Find the integers.
The answers are : -7 and 5 or 7 and -5.

Could you please show me how to work this problem out?
• February 12th 2010, 01:22 PM
Jhevon
Let the integers be $x$ and $y$
Quote:

Originally Posted by Tessarina
Two integers differ by 12

this means $x - y = 12$

Quote:

and the sum of their squares is 74.
this means $x^2 + y^2 = 74$

Quote:

Find the integers.
What can you do with the above equations?
• February 12th 2010, 02:26 PM
Tessarina
Quote:

Originally Posted by Jhevon
Let the integers be $x$ and $y$this means $x - y = 12$

this means $x^2 + y^2 = 74$

What can you do with the above equations?

That is where i got up to, but i dont know how to get to the final answer.
• February 12th 2010, 02:33 PM
icemanfan
Substitute

$x = y + 12$

into the equation $x^2 + y^2 = 74$

yielding $(y + 12)^2 + y^2 = 74$,

which can be solved using the quadratic formula.
• February 12th 2010, 07:08 PM
Tessarina
Quote:

Originally Posted by icemanfan
Substitute

$x = y + 12$

into the equation $x^2 + y^2 = 74$

yielding $(y + 12)^2 + y^2 = 74$,

which can be solved using the quadratic formula.

Could you please show me the working out to get to the answer?
• February 13th 2010, 04:19 AM
Hello Tessarina
Quote:

Originally Posted by Tessarina
Could you please show me the working out to get to the answer?

$(y+12)^2+y^2=74$

$\Rightarrow y^2+24y+144+y^2=74$

$\Rightarrow 2y^2+24y +70=0$

$\Rightarrow y^2+12y+35=0$

$\Rightarrow (y+5)(y+7)=0$

$\Rightarrow y = -5, -7$

$\Rightarrow x = y+12 = 7, 5$

So the solutions are $(7, -5)$ and $(5, -7)$.