Question

using quadratic formula, solve the following equation 9x^2-9(a+b)x+2a^2+5ab+2b^2)=0

Answer 1

Given Equation is 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0

On comparing with ax^2 + bx + c = 0, we get

a = 9, b = -9(a + b), c = 2a^2 + 5b^2 + 2b^2.

The solutions are:

(1)

$= \ \textgreater \ x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(-9(a + b)) \sqrt{(-9(a + b)^2 - 4(9)(2a^2 + 2b^2 + 5ab)} }{2 * 9}$

$= \ \textgreater \ \frac{9(a + b) \sqrt{81(a + b)^2 - 36(2a^2 + 2b^2 + 5ab)} }{18}$

$= \ \textgreater \ \frac{9(a + b) \sqrt{81(a^2 + b^2 + 2ab) - 36(2a^2 + 2b^2 + 5ab)} }{18}$

$= \ \textgreater \ \frac{9(a + b) \sqrt{81a^2 + 81b^2 + 162ab - 72a^2 - 72b^2 - 180} }{18}$

$= \ \textgreater \ \frac{9(a + b) \sqrt{9a^2 + 9b^2 - 18ab} }{18}$

$= \ \textgreater \ \frac{9(a + b) \sqrt{9(a^2 + b^2 - 2ab)} }{18}$

$= \ \textgreater \ \frac{9(a + b) \sqrt{9(a - b)^2} }{18}$

$= \ \textgreater \ \frac{9(a + b) + 3(a - b)}{18} , \frac{9(a + b) - 3(a - b)}{18}$

$= \ \textgreater \ \frac{9a + 9b + 3a - 3b}{18} , \frac{9a + 9b - 3a + 3b}{18}$

$= \ \textgreater \ \frac{12a + 6b}{18}, \frac{6a + 12b}{18}$

$= \ \textgreater \ \frac{6(2a + b)}{18} , \frac{6(a + 2b)}{18}$

$= \ \textgreater \ \frac{2a + b}{3} , \frac{a + 2b}{3}$



Hope this helps!