How do you solve 1+ 6( 2x + 3) ^- 1+ 8( 2x + 3) ^- 2= 01+6(2x+3)1+8(2x+3)2=0?

2 Answers
Sep 13, 2016

The Soln. : x=-7/2, or, x=-5/2:x=72,or,x=52

Explanation:

1+6(2x+3)^(-1)+8(2x+3)^(-2)=01+6(2x+3)1+8(2x+3)2=0

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

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

y^2+6y+8=0y2+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.!

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