1.

Solve the following quadratic equations by factorization:\(ax^{2}+(4a^{2}-3b)x-12ab = 0\)

Answer»

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(ax^{2}+(4a^{2}-3b)x-12ab=0\)

⇒ ax2 + 4a2x – 3bx – 12ab = 0 

⇒ ax(x + 4a) -3b(x + 4a) = 0 

⇒ (ax – 3b)(x + 4a) = 0 

⇒ x = 3b/a, -4a



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