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


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:


\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}


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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