Two numbers differ by 4, but their product is 96.

Question

Hi 

I need to find both numbers using factorising methods.

Thanks :)

Answer 1

Often, after learning the "standard answer," good students will want to explore other answers in hopes of finding "shortcuts" or other approaches to standard problems.  This allows students to apply new concepts to old problems.  You will find, for example, that there is a simple equation for the area of a triangle (A=bh/2), but is quite enlightening to obtain that formula in calculus class by integrating the formula y=x from 0 to b.

 

So, after you read Arturo O.'s thorough answer, consider the fact that you may assign x to (a) the lower number, (b) the higher number, or (c) the average of the numbers.

 

Method 3 is worth considering:

  Let x be a number such that the two numbers to be found are (x-2) and (x+2).

  We've learned that  (x-a)(x+a) = x² - a²

    So,  (x-2)(x+2) = x²-4

 

Since the product of the two numbers is 96,

       x² - 4 = 96

       x² = 100

       x = ±10

 

Thus, the numbers are either:

      -12 and -8     (that is -10-2 and -10+2)

   or

       8 and 12      (that is, 10-2 and 10+2)

Answer 2

n, m are the 2 numbers.

 

(i)  n - m = 4

 

(ii) nm = 96

 

From (i) get n = m + 4, plug it into (ii) to get m, then plug m into either equation to get n.

 

(m + 4)m = 96

 

m² + 4m - 96 = 0

 

Factor this equation and get

 

(m - 8)(m + 12) = 0

 

m = 8 or -12

 

For m = 8, n = m + 4 = 8 + 4 = 12

For m = -12, n = m + 4 = -12 + 4 = -8

 

There are 2 solutions:

 

n = 12 and m = 8

 

or 

 

n = -8 and m = -12

 

 

 

Answer 3

let the two numbers be x and x+4.  Now can you proceed?

 

In your question, the word but should have been and.