Question

Solve
$9{x}^{2} - 9(a + b)x + (2 {a}^{2} + 5ab + 2 {b}^{2}) = 0$
please solve with the help of quadratic roots formula

thanks and 100 points is big deal though
it is a bit crazy but we are mathematicians

Answer 1

Heya friend....☺

Here is your solution ......⭐⭐⭐⭐⭐

Hope this will help you......☺☺

Thanks......❇❇❇

Answer 2

heya...!!!!

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

a = 9

b = -9(a + b)

c = (2a² + 5ab + 2b²)

we know that:-

D = (b² - 4ac)

=> [-9(a + b)]² - 4×(9)×(2a + b)(a + 2b)

=> 81(a² + b² + 2ab) - 36(2a² + 5ab + 2b²)

=> 81a² + 81b² + 162ab - 72a² - 180ab - 72b²

=> 81a² - 72a² + 81b² - 72b² + 162ab - 180ab

=> 9a² + 9b² - 18ab

=> (3a)² + (3b)² - 2×3×3×ab

=> (3a - 3b)²

=> √D = (3a - 3b)

now,

$x = \frac{ - ( b) + - \sqrt{D}}{2a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: take(+) sign. \\ \\ \\ = > x = \frac{ - [- 9(a + b)] + (3a - 3b) }{2 \times 9} \\ \\ = > x = \frac{ + 9a + 9b + 3a - 3b}{18} \\ \\ = > x = \frac{ 6(2a + b)}{18} \\ \\ = > x = \frac{ (2 a + b)}{3} \\ \\ \\ \: \: \: \: take \: (-) sign. \\ \\ = > x = \frac{ + 9a + 9b - 3a + 3b}{18} \\ \\ = > x = \frac{ 6(a + 2b) }{18} \\ \\ = > x = \frac{ (a + 2b)}{3}$