Alright. For this problem, I tried just counting random integers and see if their product is 45 and their sum 12. Using this terrible method, I found 18=15+3. There has to be a better way...

-*\(tommarvoloriddle\)*

tommarvoloriddle Jul 2, 2019

#1**+4 **

You read it wrong!!! Oops!

It said **difference**, not sum

This is my way, there might be a better way.

When it says "their product is 45" I just find its factor pairs

5 * 9

15 * 3

Those are the first to come into my mind

Then I instantly recognize 15 and 3 has a DIFFERENCE of 12.

You can make a system of equations, but that is too slow.

CalculatorUser Jul 2, 2019

#3**+5 **

um no. I was just showing my last steps, which was 15+3=18. Because 15 and 3 were a pair...

tommarvoloriddle
Jul 2, 2019

#4**+3 **

**Help.**

\(\begin{array}{|lrcll|} \hline (1) & (a+b)^2 &=& a^2+2ab+b^2 \\ (2) & (a-b)^2 &=& a^2-2ab+b^2 \\ \hline (1)-(2): & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- (a^2-2ab+b^2) \\ & (a+b)^2-(a-b)^2 &=& a^2+2ab+b^2- a^2+2ab-b^2 \\ & (a+b)^2-(a-b)^2 &=& 4ab \\ & (a+b)^2 &=& 4ab+(a-b)^2 \\ & \mathbf{ a+b } &=& \mathbf{\sqrt{4ab+(a-b)^2}} \quad | \quad ab=45,\quad a-b = 12 \\ & a+b &=& \sqrt{4\cdot 45+12^2} \\ & a+b &=& \sqrt{324} \\ &\mathbf{ a+b} &=& \mathbf{18} \\ \hline \end{array} \)

The sum of the integers is **18**

heureka Jul 2, 2019

#5**+4 **

Thank you Heureka!

EDIT: why did you say help at the top of your answer? It just caught my eye.

-\(tommarvoloriddle\)

tommarvoloriddle
Jul 2, 2019

#6**+1 **

Let a be the larger integer and b be the smaller....so...we have

a - b = 12 and ab = 45

Rewrite the first equation as a = 12 + b

Sub this into the second for a

(12 + b) b = 45 simplify

12b + b^2 = 45 rearrange as

b^2 + 12b - 45 = 0

We need two factors that multiply to -45 and sum to 12

These are 15 and -3

So we write

(b +15) (b -3) = 0 set each factor to 0 and solve for b and we get that b = -15 (reject) and b = 3

And .... a = 12 + b = 12 + 3 = 15

And their sum is 18

CPhill Jul 2, 2019