+0  
 

Best Answer 

 #1
avatar+797 
+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.

GYanggg  Apr 16, 2018
 #1
avatar+797 
+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

 

\(ax^2+bx+c=0\)

 

the two roots are \(p\) and \(q\) then we can put \(x^2+bx+c\equiv(x-p)(x-q)\)

and therefore \(a=1,\ b=-(p+q),\ c=pq\).

It follows that \(p\) and \(q\) are the two roots of the equation \(x^2-16x+55=0\).

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

 

\(\displaystyle x = {-b \pm \sqrt{b^2-4ac} \over 2a} = {16\pm \sqrt{16^2-4\times 55}\over 2}=8\pm3\)

 

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

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

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