The roots of the equation ax^{2} + (4a^{2} - | vedclass

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Question

(B) The given quadratic equation is $ax^{2} + (4a^{2} - 3b)x - 12ab = 0$.Expanding the middle term,we get $ax^{2} + 4a^{2}x - 3bx - 12ab = 0$.Grouping

Vedclass · Accepted Answer

$-4a, \frac{3b}{a}$

Vedclass · Answer

(B) The given quadratic equation is $ax^{2} + (4a^{2} - 3b)x - 12ab = 0$.Expanding the middle term,we get $ax^{2} + 4a^{2}x - 3bx - 12ab = 0$.Grouping the terms,we have $ax(x + 4a) - 3b(x + 4a) = 0$.Factoring out $(x + 4a)$,we get $(ax - 3b)(x + 4a) = 0$.Setting each factor to zero,we get $ax - 3b = 0$ or $x + 4a = 0$.Solving for $x$,we find $x = \frac{3b}{a}$ or $x = -4a$.Thus,the roots of the equation are $-4a$ and $\frac{3b}{a}$.

The roots of the equation $ax^{2} + (4a^{2} - 3b)x - 12ab = 0$ are

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