Question

26. Find two numbers whose sum is 21 and product is 108.​

Answer 1

12 & 9 are the numbers whose product is 108 and sum is 21

Answer 2

Let one integer be x

Then the second integer = 21 - x.

Their product = 108

ie. x(21-x) = 108

21x - x^2 = 108

21x - x^2 - 108 = 0

-x^2 + 21x - 108 = 0

Multiplying the whole with -1, (in order to make the co-efficient of x^2 +ve)

x^2 - 21x + 108 = 0 -------- (1)

[Now we have to find two numbers such that their product is 108 and sum is -21. and these two numbers will eb factors of 108. They are -9 and -12]

So (1) can be written as

x^2 - 9x - 12x + 108 = 0

x(x-9) - 12(x-9) = 0

(x-9) (x-12) = 0

So either x-9 = 0 or x-12 = 0

x = 9 or x = 12

These are the required integers.

$\huge\sf\bold\red{Thanks}$