How do you solve 2/(x+1) + 5/(x-2)=-22x+1+5x2=2?

2 Answers
Feb 13, 2016

the roots are -33 and +1/2+12

Explanation:

Starting point
1) 2/(x+1) + 5/(x-2) = -22x+1+5x2=2

Multiply throughout by x+1x+1
2) 2 + (5x-5)/(x-2) = -2*(x+1)2+5x5x2=2(x+1)

Multiply throughout by (x-2)(x2)
3) 2x-4+5x-5=-2*(x+1)*(x-2)2x4+5x5=2(x+1)(x2)

Simplifying
4) 7x+1 = -2x^2+2x+47x+1=2x2+2x+4

Gathering like terms
5) 2x^2 +5x-3=02x2+5x3=0

Using the quadratic formula (-b +-sqrt(b^2-4ac))/(2a)b±b24ac2a

and substituting in the values gives
6) -5+-sqrt(5^2-(4*2*-3))/(2*2)5±52(423)22

Simplifying
7) (-5+-7)/45±74

Gives the following answers
-12/4 = -3124=3
and
2/4=+1/224=+12

Feb 14, 2016

color(green)(x=1/2,-3x=12,3

Explanation:

color(blue)(2/(x+1)+5/(x-2)=-22x+1+5x2=2

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

rarr(x+1*2/(x+1))+(x+1*5/(x-2))=-2*(x+1)(x+12x+1)+(x+15x2)=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

:)