Question
Solve the following quadratic equations by factorization:
$ a x^2+\left(4 a^2-3 b\right) x-12 a b=0$
$ a x^2+\left(4 a^2-3 b\right) x-12 a b=0$
✓
Answer
We have,
$ a x^2+\left(4 a^2-3 b\right) x-12 a b=0$
$ {\left[a \times 12 a b=-12 a^2 b^2=4 a^2 \times-3 b\right]}$
$ \Rightarrow a x^2+4 a^2 x-3 b x+(4 a \times(-3 b))=0$
$ \Rightarrow a x(x+4 a)-3 b(x+4 a)=0$
$ \Rightarrow(x+4 a)(a x-3 b)=0$
$ \Rightarrow(x+4 a)=0 \text { or }(a x-3 b)=0$
⇒ x = -4a or $\text{x}=\frac{3\text{b}}{\text{a}}$
$\therefore\text{x}=\frac{3\text{b}}{\text{a}}$ and x = -4a are the two roots of the given equations.
$ a x^2+\left(4 a^2-3 b\right) x-12 a b=0$
$ {\left[a \times 12 a b=-12 a^2 b^2=4 a^2 \times-3 b\right]}$
$ \Rightarrow a x^2+4 a^2 x-3 b x+(4 a \times(-3 b))=0$
$ \Rightarrow a x(x+4 a)-3 b(x+4 a)=0$
$ \Rightarrow(x+4 a)(a x-3 b)=0$
$ \Rightarrow(x+4 a)=0 \text { or }(a x-3 b)=0$
⇒ x = -4a or $\text{x}=\frac{3\text{b}}{\text{a}}$
$\therefore\text{x}=\frac{3\text{b}}{\text{a}}$ and x = -4a are the two roots of the given equations.
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