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Question
ax2 + (4a2 - 3b)x - 12 ab = 0
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Solution
ax2 + (4a2 - 3b)x - 12 ab = 0
⇒ ax2 + 4a2x - 3bx - 12ab = 0
⇒ ax (x + 4a) - 3b(x + 4a) = 0
⇒ (ax - 3b) (x + 4a) = 0
⇒ x = `(3b)/a or -4a` are two roots of equation.
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