Question

If a - b = 4 and a + b = 6; find
(i) a² + b²
(ii) ab

Answer 1

(i) We know that,
( a - b )² = a² - 2ab + b² 
Rewrite the above identity as,
a²  + b² = ( a - b )² + 2ab           ....(1)
Similarly, we know that,
( a + b )² = a² + 2ab + b²
Rewrite the above identity as,
 a²  + b² = ( a + b )² - 2ab                                     .....(2)
Adding the equations (1) and (2), we have
2( a² + b² ) = ( a - b )² + 2ab + ( a + b )² - 2ab
⇒ 2( a² + b² ) = ( a - b )²  + ( a + b )²
⇒ ( a² + b² ) = `1/2[( a - b )^2  + ( a + b )^2]`          ....(3)

Given that a + b = 6 ; a - b = 4
Substitute the values of ( a + b ) and (a - b)
in equation (3), we have
a² + b² = `1/2[ (4)^2 + (6)^2]`

= `1/2[ 16 + 36 ]`

= `52/2`
⇒ ( a² + b² ) = 26                                            .....(4)

From equation (4), we have
a² + b² = 26
Consider the identity,
( a - b )² = a² + b² - 2ab                                ....(5)
Substitute the value a - b = 4 and a² + b² = 26
in the above equation, we have
(4)² = 26 - 2ab
⇒ 2ab = 26 - 16
⇒  2ab = 10
⇒  ab = 5