Question

How do you solve `1+ 6( 2x + 3) ^- 1+ 8( 2x + 3) ^- 2= 0`?

Answer 1

The Soln. `: x=-7/2, or, x=-5/2`

Explanation:

`1+6(2x+3)^(-1)+8(2x+3)^(-2)=0`

Multiplying by, `(2x+3)^2`, we get,

`(2x+3)^2+6(2x+3)+8=0`. Letting, `(2x+3)=y`, we get,

`y^2+6y+8=0`

`:. (y+4)(y+2)=0`

`:. y=-4, or, y=-2`

`:. 2x+3=-4, or, 2x+3=-2`

`:. 2x=-7, or, 2x=-5`

`:. x=-7/2, or, x=-5/2`

These roots satisfy the given eqn.

Hence, the Soln. `: x=-7/2, or, x=-5/2`

Spread the Joy of Maths.!

Answer 2

`x = -7/2 = -3 1/2 " or " x = -5/2 = -2 1/2`

Explanation:

Recall: `x^-m = 1/x^m`

Get rid of the negative indices first

`1+ 6( 2x + 3) ^- 1+ 8( 2x + 3) ^- 2= 0`

`1+ 6/( 2x + 3) + 8/( 2x + 3) ^2= 0`

As it is an equation, we can get rid of the denominators by multiplying through by the LCD which in this case is `color(red)((2x+3)^2)`

`1xxcolor(red)((2x+3)^2)+ (6xxcolor(red)((2x+3)^2))/( 2x + 3) + (8xxcolor(red)((2x+3)^2))/( 2x + 3) ^2= 0`

`color(red)((2x+3)^2)+ 6color(red)((2x+3)) + 8= 0`

`4x^2 +12x+9 +12x +18 +8 = 0`

`4x^2+24x +35 =0 " " larr` find factors

`(2x+7)(2x+5)=0`

If `2x+7 = 0 rarr x = -7/2`

If `2x+5 = 0 rarr x = -5/2`