Question

The equation $\bf \dfrac{24 {x}^{2} + 25x - 47 }{ax-2} = - 8x - 3 - \dfrac{53}{ax - 2}$ is true for all values of $\bf x \neq \dfrac{2}{a}$ , where a is a constant.

What is the value of a?

(A) -16

(B) -3

(C) 3

(D) 16​

Answer 1

EXPLANATION.

equation are

=>

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

conditions are = x ≠ 2/a , where a is

constant.

TO FIND VALUE OF A.

multiply both sides of equation by ( ax - 2 )

we get,

=> 24x² + 25x - 47 = ( ax - 2 ) ( -8x - 3 ) - 53

=> 24x² + 25x + 6 = ( ax - 2 ) ( -8x - 3 )

=> 24x² + 25x + 6 = - 8ax² - 3ax + 16x + 6

=> 24x² + 9x = -8ax² - 3ax

=> 24x² + 8ax² + 9x + 3ax = 0

=> 8x² ( 3 + a) + 3x ( 3 + a) = 0

=> ( 8x² + 3x ) ( a + 3 ) = 0

=> a = -3

for the given conditions

=> x ≠ 2/a

=> x ≠ -2/3

Therefore,

value of a = -3

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

The equation  $\sf{\dfrac{24 \mathrm{x}^{2}+25 \mathrm{x}-47}{\mathrm{ax}-2}=-8 \mathrm{x}-3-\dfrac{53}{\mathrm{ax}-2}}$ is true for all values of $\sf{x \neq \dfrac{2}{a}}$ , where a is a constant.

♣ ɢɪᴠᴇɴ :

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

♣ ᴛᴏ ꜰɪɴᴅ :

Value of a

♣ ᴀɴꜱᴡᴇʀ :

$\large{\boxed{\sf{a=-3}}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\boxed{\underline{\underline{\sf{\dfrac{24 \mathrm{x}^{2}+25 \mathrm{x}-47}{\mathrm{ax}-2}=-8 \mathrm{x}-3-\dfrac{53}{\mathrm{ax}-2}}}}}$

$\sf{\Rightarrow \dfrac{24 \mathrm{x}^{2}+25 \mathrm{x}-47}{\mathrm{ax}-2}=\dfrac{(-8 \mathrm{x}-3)(\mathrm{ax}-2)-53}{{ax}-2}}$

$\sf{\Rightarrow 24 \mathrm{x}^{2}+25 \mathrm{x}-47=(-8 \mathrm{x}-3)(\mathrm{ax}-2)-53}$

$\sf{\Rightarrow 24 \mathrm{x}^{2}+25 \mathrm{x}-47=-8 \mathrm{ax}^{2}+16 \mathrm{x}-3 \mathrm{ax}+6-53}$

$\sf{\Rightarrow 24 \mathrm{x}^{2}+25 \mathrm{x}-47=-8 \mathrm{ax}^{2}+16 \mathrm{x}-3 \mathrm{ax}-47}$

$\sf{\Rightarrow 24 \mathrm{x}^{2}+25 \mathrm{x}=-8 \mathrm{ax}^{2}+(16-3 \mathrm{a}) \mathrm{x}}$

$\sf{\Rightarrow 24=-8 \mathrm{a}}$

$\large{\boxed{\bigstar\:\:\sf{a=-3}}}$