If 2a² - 2ab + b² + 6a + 9 = 0, then what is the value of (a+b) + ab?

Dear Student,

Please find below the solution to the asked query:

We have :  2 a² - 2 ab + b² + 6 a + 9 = 0  , So

$\Rightarrow$ a² - 2 ab + b² + a² + 6 a + 9 = 0

$\Rightarrow$( a² - 2 ab + b² ) + ( a² + 2 ( a ) ( 3 ) + 3² )= 0

$\Rightarrow$( a - b )² + ( a + 3 )²  = 0

We know sum of any two square term only be equal to zero if all terms ( here both terms ) are equal to zero , So

( a - b )² = 0

a - b = 0

a = b                                                      --- ( 1 )

And

( a + 3 )² = 0 ,

a + 3 = 0

a = - 3 , From equation 1 we get

a =  b = - 3

So ,

Values of ( a + b ) + a b we get by substituting values of ' a ' and ' b ' . 

$\Rightarrow$ ( - 3 + ( - 3 ) ) + ( - 3 ) ( - 3 )

$\Rightarrow$ ( - 3 - 3 ) + 9

$\Rightarrow$ - 6 + 9

$\Rightarrow$ 3                                                                                                              ( Ans )

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