Question

✬ Hey There ✬

$\frac{24x ^{2}+ 25x - 47 }{ax - 2} = - 8x - 3 - \frac{53}{ax - 2}$
The above equation is true for all values of x ≠ 2/a , where a is constant. So what will be the value of a ?

Options are

A) -3
B) -16
C) 16
D) 3​

Answer 1

EXPLANATION.

$\sf \: \implies \dfrac{24 {x}^{2} + 25x - 47 }{ax - 2} = - 8x - 3 - \dfrac{53}{ax - 2} \\ \\ \sf \implies \: \: a \: \ne \: \frac{2}{a}$

$\sf \: \implies \: \dfrac{24 {x}^{2} + 25x - 47}{ax - 2} = \dfrac{( - 8x - 3)(ax - 2) - 53}{ax - 2} \\ \\ \sf \: \implies \: 24 {x}^{2} + 25x - 47 = ( - 8x - 3)(ax - 2) - 53 \\ \\ \sf \: \implies \: 24 {x}^{2} + 25x - 47 = - 8a {x}^{2} + 16x - 3ax + 6 - 53 \\ \\ \sf \: \implies \: 24 {x}^{2} + 25x - 47 = - 8 {x}^{2} + 16x - 3ax - 47$

$\sf \: \implies \: 24 {x}^{2} + 25x = - 8a {x}^{2} +16x - 3ax \\ \\ \sf \: \implies \: 24 {x}^{2} + 9x = - 8a {x}^{2} - 3ax \\ \\ \sf \: \implies \: 24 {x}^{2} + 9x + 8a {x}^{2} + 3ax = 0 \\ \\ \sf \: \implies \: 8(3 + a) {x}^{2} + 3(3 + a)x = 0 \\ \\ \sf \: \implies \: a \: = - 3$

Answer 2

$\sf \underline {Given\: equation}:\\\\\sf \frac{24x^2+25x-47}{ax-2} = -8x-3 -\frac{53}{ax-2} \\\\\\ \sf \underline {Simplifying \: RHS \: first}:\\\\\\ \implies \frac{24x^2+25x-47}{ax-2} = -8x-3 -\frac{53}{ax-2} \\\\\\ \implies \frac{24x^2+25x-47}{ax-2} = -8x-3 \frac{(ax-2)}{(ax-2)} - \frac{53}{ax-2} \\\\\\ \implies \frac{24x^2+25x-47}{ax-2} = \frac{-8x-3 (ax-2)}{(ax-2)} - \frac{53}{ax-2} \\\\\\ \implies \frac{24x^2+25x-47}{ax-2} = \frac{-8ax^2+16x-3ax+6-53}{ax-2}$

$\sf \implies 24x^{2}+25x-47 =-8ax^{2}+16x-3ax-47 \\\\ \implies 24x^2+25x = -8ax^2+16x-3ax \\\\ \implies 24x^{2}+8ax^{2}+25x-16x+3ax=0\\\\ \sf \implies 24x^2+8ax^2+9x+3ax=0\\\\\sf \underline {Taking \: x \: in \: common}:\\\\\sf \implies x(24x+8ax+9+3a)=0 \\\\ \sf \underline {Factorizing}:\\\\ \implies x \bigg(8x(3+a)+3(3+a) \bigg)=0$

$\sf \underline {Taking \: (3+a) \:in \: common}:\\\\ \implies x((8x+3)(3+a))=0\\\\\sf \underline {Dividing \: both \: sides \: by \: x, \: to\: remove \: the \: variable \: x}:\\\\ \implies (8x+3)(3+a)=0\\\\\sf \underline {Solving \: for \: a}:\\\\ \implies 3+a = 0\\ \implies a = -3$