Question

If the sum of two numbers is 13 and their product is 42. Determine the nember.

Answer 1

Answer:

6 and 7 are the required numbers.

Step-by-step explanation:

Let the numbers be a and b. ( b < a )

Given,

Sum of numbers = 13 = a + b

Product of numbers = 42 = ab

Method 1 :

= > a + b = 13

= > a = 13 - b ...( 1 )

= > ab = 42

= > ( 13 - b )b = 42 { from ( 1 ) }

= > 13b - b^2 = 42

= > b^2 - 13b + 42 = 0

= > b^2 - ( 7 + 6 )b + 42 = 0

= > b^2 - 7b - 6b + 42 = 0

= > b( b - 7 ) - 6( b - 7 ) = 0

= > ( b - 7 )( b - 6 ) = 0

Since their product is 0, one of them must be 0.

If b - 6 = 0, b = 6 & a = 13 - b = 13 - 6 = 7

If b - 7 = 0, b = 7 & a = 13 - b = 13 - 7 = 6

In both of the above cases, numbers are 6 and 7.

Thus, 6 and 7 are the required numbers.

Method : 2

= > a + b = 13

= > ( a + b )^2 = 13^2

= > a^2 + b^2 + 2ab = 169

= > a^2 + b^2 + 2ab - 4ab = 169 - 4ab { adding - 4ab to both sides }

= > a^2 + b^2 - 2ab = 169 - 4( 42 ) { ab = 42 }

= > ( a - b )^2 = 169 - 168 = 1 { a^2 - 2ab + b^2 = ( a - b )^2 }

= > a - b = 1

Now, a + b = 13 & a - b = 1

= > ( a + b ) + ( a - b ) = 13 + 1 { adding both }

= > 2a = 14 = > a = 7 = > b = 13 - a = 13 - 7 = 6.

Hence, numbers are 6 and 7.

Answer 2

Answer:

=> a+b=13

=>(a+b)^2 = 13^2

=>a^2+b^2+2ab = 169

=>a^2+b^2+2ab - 4ab = 169 - 4ab

4ab to both sides)

=>a^2+b^2-2ab169-4(42)

(adding-

(ab=42)

=>(a - b)^2 = 169-168=1

{a^2-2ab+b^2=

(a - b)^2)

=>a-b=1

Now, a + b = 13 & a-b=1

=>(a+b)+(a - b) = 13+1 { adding both} => 2a = 14 =>a=7=> b=13-a 13-7= 6.

Hence, numbers are 6 and 7.