Maths-

General

Easy

Question

# Consider the difference of squares π^{2} β π^{2} for integers π and π. Make a table for the difference of squares using consecutive integers for π and π. What pattern do you notice? Using the pattern find the pair of consecutive integers that generate the difference of squares of β45.

Hint:

### The methods used to find the product of binomials are called special products.

Difference of squares is a case of a special product which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign

## The correct answer is: 22 and 23

### a and b are consecutive integers. Letβs make a table for some pairs of a and b

We can notice that the difference of square of consecutive integers a and b is of the form -(a + b).

If the difference of two consecutive numbers is -45. We can think that there is only one case when a and b in -(a + b) is connective i.e. when a = 22 and b = 23.

Final Answer:

Hence, the pattern for difference of squares using consecutive integers for π and π is -(a + b) and the pair of consecutive integers that generate the difference of squares of β45 is 22 and 23

We can notice that the difference of square of consecutive integers a and b is of the form -(a + b).

If the difference of two consecutive numbers is -45. We can think that there is only one case when a and b in -(a + b) is connective i.e. when a = 22 and b = 23.

Final Answer:

Hence, the pattern for difference of squares using consecutive integers for π and π is -(a + b) and the pair of consecutive integers that generate the difference of squares of β45 is 22 and 23