How do you solve #2/(x+1) + 5/(x-2)=-2#?

2 Answers
Feb 13, 2016

Answer:

the roots are #-3# and #+1/2#

Explanation:

Starting point
1) #2/(x+1) + 5/(x-2) = -2#

Multiply throughout by #x+1#
2) #2 + (5x-5)/(x-2) = -2*(x+1)#

Multiply throughout by #(x-2)#
3) #2x-4+5x-5=-2*(x+1)*(x-2)#

Simplifying
4) #7x+1 = -2x^2+2x+4#

Gathering like terms
5) #2x^2 +5x-3=0#

Using the quadratic formula #(-b +-sqrt(b^2-4ac))/(2a)#

and substituting in the values gives
6) #-5+-sqrt(5^2-(4*2*-3))/(2*2)#

Simplifying
7) #(-5+-7)/4#

Gives the following answers
#-12/4 = -3#
and
#2/4=+1/2#

Feb 14, 2016

Answer:

#color(green)(x=1/2,-3#

Explanation:

#color(blue)(2/(x+1)+5/(x-2)=-2#

Multiply everything with #x+1# to get rid of the denominator:

#rarr(x+1*2/(x+1))+(x+1*5/(x-2))=-2*(x+1)#

#rarr(cancel(x+1)2/cancel(x+1))+((5*(x+1))/(x-2))=-2*(x+1)#

Remove the brackets:

#rarr2+(5*(x+1))/(x-2)=-2*(x+1)#

Use distributive property #color(orange)(a(b+c)=ab+ac#

#rarr2+(5x+5)/(x-2)=-2x-2#

Add #2# both sides:

#rarr4+(5x+5)/(x-2)=-2x#

Multiply everything with #x-2# to get rid of the denominator:

#rarr4*(x-2)+(x-2*(5x+5)/(x-2))=-2x*(x-2)#

#rarr4x-8+(cancel(x-2)(5x+5)/cancel(x-2))=-2x^2+4x#

Remove brackets:

#rarr4x-8+5x+5=-2x^2+4x#

Subtract #4x# both sides:

#rarr-8+5x+5=-2x^2#

#rarr5x-3=-2x^2#

Add #-2x^2# both sides:

#rarr2x^2+5x-3=0#

Now this is a Quadratic equation (in form #ax^2+bx+c=0#)

Use Quadratic formula:

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case,

#color(red)(a=2,b=5,c=-3#

Substitute the values:

#rarrx=color(brown)((-(5)+-sqrt(5^2-4(2)(-3)))/(2(2))#

#rarrx=color(brown)((-5+-sqrt(25-(-24)))/4#

#rarrx=color(brown)((-5+-sqrt(25+24))/4#

#rarrx=color(brown)((-5+-sqrt(49))/4#

#rarrx=color(indigo)((-5+-7)/4#

So, now #x# has #2# values:

#rarrcolor(blue)x=color(violet)((-5+7)/4,(-5-7)/4#

So we can first solve for the first value:

#rarrx=(-5+7)/4=2/4=color(green)(1/2#

Now,for the second value:

#rarrx=(-5-7)/4=-12/4=color(green)(-3#

:)