Question

Find the roots of the given quadratic equation using factorization. 6x2+(a 12)x (a2 + a-6) 0 (a 0).

Answer 1

Answer:

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Answer 2

The roots of the given quadratic equation $6x^2+(a - 12)x - (a^2 + a-6) = 0$, obtained using the factorization method are $x = \frac{2-a}{2}$ and $x = \frac{a+3}{3}$.

Given:

$6x^2+(a - 12)x - (a^2 + a-6) = 0$ where, $a \neq 0.$

To Find:

We have to find out the roots of the given quadratic equation using factorization.

Solution:

The given quadratic equation is,

$6x^2+(a - 12)x - (a^2 + a-6) = 0$

Let $q_{1} + q_{2} = a-12$ and $q_{1} . q_{2} = (-6).(a^2+ a-6)$.

Now consider the term $(-6).(a^2+ a-6)$. This term can be factorized as follows.

$(-6).(a^2+ a-6)$ ⇒ $(-6).[(a+3)( a-2)]$

⇒ $(-2).(a+3).3. ( a-2)$

⇒$-(2a+6)(3a-6)$

⇒$(3a-6) (-2a-6)$.

$i.e., q_{1} = (3a-6)$ and $q_{2} = (-2a-6)$.

∴, The given quadratic equation becomes,

$6x^2+[(3a - 6) - (2a + 6)]x - (a^2+a-6) = 0$

$6x^2+(3a - 6)x - (2a + 6)x - (a+3)(a-2) = 0$

$3x [2x+(a-2)]- (a+3) [2x+(a-2)] = 0$

$[2x+(a-2)] [3x-(a+3)] = 0$

Consider  $[2x+(a-2)] = 0$

∴, $2x = 2-a$

$i.e., x = \frac{2-a}{2}$.

Similarly, consider $3x-(a+3) = 0$

∴, $3x = a+3$

$i.e., x = \frac{a+3}{3}$.

Hence, the roots of the given quadratic equation using factorization are $x = \frac{2-a}{2}$ and $x = \frac{a+3}{3}$.

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