Question

find two consecutive numbers whose sum is 27 and product is 182​

Answer 1

C O N C E P T :

♣ First assume any two variables for the 2 required numbers. Now by using the first condition find a relation between these 2. Then by using the second condition find another relation between two variables. By using these 2 relations find their values. The values of these variables are the required result.

S O L U T I O N :

♠ Given conditions in the question, can be written here as :

♠The Sum of two random numbers is given to be 27.

Product of above two numbers is given to be as 182.

♠ Let us assume the first number to be denoted by x.

♠ Let us assume the second number to be denoted by y.

♠ By writing the first condition, mathematically, we get it as :

: $\implies$ x + y = 27 ( i )

By writing the second condition, mathematically, we get it as :

: $\implies$ xy = 182. ( ii )

By Subtracting y on both sides of equation (i), we get it as :

: $\implies$ x + y - y = 27 - y

By cancelling common term, we can write the equations in form of :

: $\implies$ x = 27 - y ( iii )

By Substituting equation (iii) into the equation (ii), we get it as .

: $\implies$ ( 27 - y ) y = 182

Distributive law is defined as product of 2 terms, given by :

: $\implies$ a ( b + c ) = ab + ac

By using above law we can write the above equation as :

: $\implies$ 27 y - y² = 182

By adding y² and subtracting 27 y on both sides, we get

: $\implies$ y² - 27 y + 182 = 0

Factorizations : For factoring a quadratic follow the steps :

Allot x² coefficient as "a", x co-efficient as "b", Constant as "c”.

Find the product of the 2 numbers a ,c. Let it be K.

Find 2 numbers such that their product is K, Sum is b.

Rewrite the term bx

in terms of those 2 numbers.

Now factor the first two terms.

Next factor the last two terms.

If we do it correctly our 2 new terms will have one more common factor.

Now take that common to get quadratic as a product of 2 terms, for which you get the roots.

By comparing it to ax² + bx + c = 0, we get values of a, b, c, as :

: $\implies$ a = 1, b = -27, c = 182.

Two numbers whose product is 182, sum is 27, are -13, -14.

By writing -27y as -13y-14y, we get the equation as:

: $\implies$ y² - 13 y - 14 y + 182 = 0

By the above equation, we can say roots of y are given by.

: $\implies$ y = 13, 14.

By substituting y = 13 in equation (iii), we get x as :

: $\implies$ x = 27 - 13 = 14

By substituting y = 14, in equation (iii), we get x as :

: $\implies$ x = 27 - 14 = 13

So, the 2 roots satisfying the condition all given by :

(13, 14); (14,13). Therefore these are solutions.

T I P S :

♠ The Alternate method is to divide with x on both sides of equation (ii), then substitute y in terms of x into equation (i), any ways you get the same quadratic in y. Also you can write -27y as -14y-13y instead of -13y-14y any ways you will have the same roots for y.