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

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Q) Solve the following quadratic equation by factorization method: $9x^2-9 (a+b)x +(2a^2+5ab+2b^2)=0$

Ans:

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

The Constant term $2a^{2}+5ab+2b^{2}$ can also be presented as:

$2a^{2}+4ab+ab+2b^{2}$

$\implies 2a(a+2b) + b(a + 2b)$

$\implies (2a + b) (a + 2b)$

Similarly, Coefficient of the middle term $-9(a+b)$ can be presented as:

$(-9a - 9b)$

$\implies (-6a - 3a - 3b - 6b)$

$\implies (-6a - 3b - 3a - 6b)$

$\implies -3(2a + b) - 3 (a +2b)$

$\implies -3\{(2a+b)+(a+2b)\}$

Now, the original quadratic equation:

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

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

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

$\implies 3x\{3x-(2a+b)\}-(a+2b)\{3x-(2a+b)\}=0$

$\implies \{3x-(2a+b)\}\{3x-(a+2b)\}=0$

$\implies \{3x-(2a+b)\}=0$

Or, $\{3x-(a+2b)\}=0$

Now, if $\{3x-(2a+b)\}=0 \implies x=\frac{2a+b}{3}$

and

if $\{3x-(a+2b)\}=0 \implies x=\frac{a+2b}{3}$

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

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