1. ## Problem solving with Quadratics

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?

2. Let the integers be $\displaystyle x$ and $\displaystyle y$
Originally Posted by Tessarina
Two integers differ by 12
this means $\displaystyle x - y = 12$

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

Find the integers.
What can you do with the above equations?

3. 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.

4. 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.

5. 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?

6. Hello Tessarina
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 $\displaystyle (5, -7)$.