Question

what will be the answer for this​

Answer 1

Given:

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

To Find:

Find the value of x from the equation

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

Solution:

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

or, $9x^{2} - 9(a+b)x + [2a^{2} + (4+1)ab +2b^{2}] = 0$

or, $9x^{2} - 9(a+b)x + [2a^{2} + 4ab+ab +2b^{2}] = 0$

or, $9x^{2} - 9(a+b)x + [2a(a + 2b)+b(a +2b)] = 0$

or, $9x^{2} - 9(a+b)x + (a + 2b)(2a+b) = 0$

or, $9x^{2} - 3[(a + 2b)+(2a+b)]x + (a + 2b)(2a+b) = 0$

or, $9x^{2} - 3(a + 2b)x+3(2a+b)x + (a + 2b)(2a+b) = 0$

or, $3x[3x - (a + 2b)]-(2a+b)[3x - (a + 2b)] = 0$

or, $(3x-a- 2b)(3x-2a-b)=0$

Either, $(3x-a- 2b)=0$

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

Or, $(3x-2a-b)=0$

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

Therefore, the values of x are $\frac{a+2b}{3}$ and $\frac{2a+b}{3}$.