Question

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

Answer 1

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`

Answer 2

`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`

:)