Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 

Best Answer 

 #1
avatar+985 
+3

The two factors of the first number that you are looking for is 11 and 5.

 

We first prime factorize 55, and we get 5 * 11.

 

So we can immediately test 5 and 11.

 

5 * 11 = 55 and 5 + 11 = 16.

 

it works!

 

i hope this helped.

 Apr 16, 2018
 #1
avatar+985 
+3
Best Answer

The two factors of the first number that you are looking for is 11 and 5.

 

We first prime factorize 55, and we get 5 * 11.

 

So we can immediately test 5 and 11.

 

5 * 11 = 55 and 5 + 11 = 16.

 

it works!

 

i hope this helped.

GYanggg Apr 16, 2018
 #2
avatar
0

I don't think this was a very good question, because you can solve it even if the two numbers aren't integer factors of 55!

The idea is to find two numbers p and q such that

 

p+q=16

and

pq=55

 

The trick is to observe that if in the quadratic equation

 

ax2+bx+c=0

 

the two roots are p and q then we can put x2+bx+c(xp)(xq)

and therefore a=1, b=(p+q), c=pq.

It follows that p and q are the two roots of the equation x216x+55=0.

You can solve this simply by "spotting" the roots or by turning the handle of the sausage machine:

 

x=b±b24ac2a=16±1624×552=8±3

 

Obviously this will work even if  b doesn't have integer factors!

 Apr 17, 2018
edited by Guest  Apr 17, 2018
edited by Guest  Apr 17, 2018

1 Online Users