Question

⇒ Find the value of a
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Answer 1

Step-by-step explanation:

We have,

$\frac{24 {x}^{2} + 25x - 47}{ax - 2} = - 8x - 3 - \frac{53}{ax - 2}$

$\implies\frac{24 {x}^{2} + 25x - 47}{ax - 2} = - ( \frac{(8x + 3)(ax - 2) + 53}{ax - 2} )$

Since, $x ≠ \frac{2}{a}$, so ax - 2 can be cancelled in both sides

$\implies24 {x}^{2} + 25x - 47 = - (8x + 3)(ax - 2) - 53$

$\implies24 {x}^{2} + 25x - 47 = - (8a {x}^{2} + 3ax - 16x - 6) - 53$

$\implies24 {x}^{2} + 25x - 47 = - 8a {x}^{2} - (3a - 16)x + 6 - 53$

$\implies24 {x}^{2} + 25x - 47 = - 8a {x}^{2} + (16 - 3a)x - 47$

On comparing both sides, we get, $a = - 3$

Answer 2

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:\dfrac{24x^2+25x-47}{ax-2}=-8x-3-\dfrac{53}{ax-2}$

To Find:

• Value of a.

Solution:

$:\longrightarrow\dfrac{24x^2+25x-47}{ax-2}=-8x-3-\dfrac{53}{ax-2}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{24x^2+25x-47}{ax-2}=\dfrac{-8x(ax-2)-3(ax-2)-53}{ax-2}\\\\:\implies\dfrac{24x^2+25x-47}{ax-2}=\dfrac{-8ax^2+16x-3ax+6-53}{ax-2}\\\\\text{Cancelling (ax-2) on both the fractions' denominators.}\\\\:\implies 24x^2+25x-47= -8ax^2+16x-3ax+6-53\\\\:\implies 24x^2+25x-47= -8ax^2+16x-3ax-47\\\\:\implies 24x^2+25x-47= -8ax^2+(16-3a)x-47 \\\\\text{As both the equations are quadratic,}\\\\\text{We compare the coefficients of x ^2 and x}\\\\:\implies24=(-8a) \:\:\:\:AND\:\:\:\:25=(16-3a)\\\\\text{So,}\\\\:\implies a=-3\:\:\:\:AND\:\:\:\:25-16=-3a\\\\:\implies a=-3\:\:\:\:AND\:\:\:\:9=-3a\\\\:\implies a=-3\:\:\:\:AND\:\:\:\:a=-3\\\\\text{\bf{So, a = -3(option B)}}$

$\text{\bf{Hence, value of a is -3.}}$