Question

solve by factorization method :9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0

Answer 1

Hey friend, Harish here

Here is your answer:

Given ,

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

Objective,

To solve it through factorization.

Solution,

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

→  $9x^{2} - 9(a + b)x + (2a^{2}+ 4ab + ab + 2b^{2}) = 0$

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

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

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

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

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

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

So,If the equation need to be zero,

Then,

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

$3x = (a+2b)$

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

Or,

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

$3x = 2a+b$

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

Hence these are the values of x.
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Hope my answer is helpful to you.

Answer 2

Hope the below answer helps u all