Question

$9 {x}^{2} - 9(a + b)x + (2 {a}^{2} + 5ab + 2 {b}^{2} ) = 0$
​

Answer 1

We have

$\rm 9 {x}^{2} - 9(a + b)x + (2 {a}^{2} + 5ab + 2 {b}^{2} ) = 0 \\ \rm here \: a = 9 \\ \rm \: b = 9(a + b) \\ \rm \:c = (2 {a}^{2} + 5ab + 2 {b}^{2})$

First we find discriminant (d) as we have to find the roots are real or not if not your question is become short this made question more efficient

Discriminant

$\rm \: d = {b}^{2} - 4ac \\ \implies \: d = ({ ( {9}^{2} (a + b)}^{2} - 4 \times 9(2 {a}^{2} + 5ab + 2 {b}^{2} )) \\ \rm \implies \: d = 9( {9a}^{2} + {9b}^{2} + 18ab - 8 {a}^{2} - 20ab - 8 {b}^{2} \\ \rm \implies \: d = 9( {a}^{2} + {b}^{2} - 2ab) \\ \rm \implies \boxed{ \: d = 9 {(a - b)}^{2}}$

Now d is positive thats why roots are real and exist so we find the question

using quadratic formula

$\rm x = \frac{ - b \pm \sqrt{d} }{2a} \\ \rm \implies x = \frac{ - ( - 9(a + b) \pm \sqrt{9 {(a - b)}^{2} } }{2 \times 9} \\ \rm \implies x = \frac{9(a + b) \pm3(a - b)}{2 \times 9} \\ \implies \rm x = \frac{3(a + b) \pm(a - b)}{6} \\ \rm \implies x = \frac{3a + 3b + a - b}{6} or \: x = \frac{3a + 3b - a + b}{6} \\ \rm \implies \:x = \frac{4a + 2b}{6}or \: x = \frac{2a + 4b}{6} \\ \rm \implies \boxed{x = \frac{2a + b}{6} or \: x = \frac{a + 2b}{6}}are \: the \: required \: solution$

Answer 2

Answer:

no no rakshit , i am sorry ! ! koi baat ka bura nhi lga...

my younger sister blocked you , she is young so she took my phn and block you ! i m sorry !

and i didn't report your answers

i will unblock you..

okay!