• Feb 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?
• Feb 12th 2010, 01:22 PM
Jhevon
Let the integers be \$\displaystyle x\$ and \$\displaystyle y\$
Quote:

Originally Posted by Tessarina
Two integers differ by 12

this means \$\displaystyle x - y = 12\$

Quote:

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

Quote:

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

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

this means \$\displaystyle 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.
• Feb 12th 2010, 02:33 PM
icemanfan
Substitute

\$\displaystyle x = y + 12\$

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

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

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

Originally Posted by icemanfan
Substitute

\$\displaystyle x = y + 12\$

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

yielding \$\displaystyle (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?
• Feb 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?

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

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

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

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

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

\$\displaystyle \Rightarrow y = -5, -7\$

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

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