+0

# For the following two numbers, find two factors of the first number such that their product is the first number and their sum is the second

0
376
2

55,16

Apr 16, 2018

#1
+974
+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
+974
+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
#2
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!

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