+0  
 
+1
229
7
avatar+1692 

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...

 

-\(tommarvoloriddle\)

 Jul 2, 2019
edited by tommarvoloriddle  Jul 2, 2019
 #1
avatar+2557 
+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.

 Jul 2, 2019
 #3
avatar+1692 
+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
 #2
avatar+2 
+1

the difference must be 15 - 3 = 12. All steps you do are correct except the last one (18 = 15 + 3)
bullet force

 Jul 2, 2019
 #4
avatar+23872 
+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

 

laugh

 Jul 2, 2019
 #5
avatar+1692 
+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
edited by tommarvoloriddle  Jul 2, 2019
 #6
avatar+106535 
+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

 

 

 

cool cool cool

 Jul 2, 2019
edited by CPhill  Jul 2, 2019
 #7
avatar+1692 
+4

Nioce CPHILL

tommarvoloriddle  Jul 2, 2019

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