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

Feb 13, 2016

the roots are $- 3$ and $+ \frac{1}{2}$

#### Explanation:

Starting point
1) $\frac{2}{x + 1} + \frac{5}{x - 2} = - 2$

Multiply throughout by $x + 1$
2) $2 + \frac{5 x - 5}{x - 2} = - 2 \cdot \left(x + 1\right)$

Multiply throughout by $\left(x - 2\right)$
3) $2 x - 4 + 5 x - 5 = - 2 \cdot \left(x + 1\right) \cdot \left(x - 2\right)$

Simplifying
4) $7 x + 1 = - 2 {x}^{2} + 2 x + 4$

Gathering like terms
5) $2 {x}^{2} + 5 x - 3 = 0$

Using the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

and substituting in the values gives
6) $- 5 \pm \frac{\sqrt{{5}^{2} - \left(4 \cdot 2 \cdot - 3\right)}}{2 \cdot 2}$

Simplifying
7) $\frac{- 5 \pm 7}{4}$

$- \frac{12}{4} = - 3$
and
$\frac{2}{4} = + \frac{1}{2}$

Feb 14, 2016

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:

$\rightarrow \left(x + 1 \cdot \frac{2}{x + 1}\right) + \left(x + 1 \cdot \frac{5}{x - 2}\right) = - 2 \cdot \left(x + 1\right)$

$\rightarrow \left(\cancel{x + 1} \frac{2}{\cancel{x + 1}}\right) + \left(\frac{5 \cdot \left(x + 1\right)}{x - 2}\right) = - 2 \cdot \left(x + 1\right)$

Remove the brackets:

$\rightarrow 2 + \frac{5 \cdot \left(x + 1\right)}{x - 2} = - 2 \cdot \left(x + 1\right)$

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

$\rightarrow 2 + \frac{5 x + 5}{x - 2} = - 2 x - 2$

Add $2$ both sides:

$\rightarrow 4 + \frac{5 x + 5}{x - 2} = - 2 x$

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

$\rightarrow 4 \cdot \left(x - 2\right) + \left(x - 2 \cdot \frac{5 x + 5}{x - 2}\right) = - 2 x \cdot \left(x - 2\right)$

$\rightarrow 4 x - 8 + \left(\cancel{x - 2} \frac{5 x + 5}{\cancel{x - 2}}\right) = - 2 {x}^{2} + 4 x$

Remove brackets:

$\rightarrow 4 x - 8 + 5 x + 5 = - 2 {x}^{2} + 4 x$

Subtract $4 x$ both sides:

$\rightarrow - 8 + 5 x + 5 = - 2 {x}^{2}$

$\rightarrow 5 x - 3 = - 2 {x}^{2}$

Add $- 2 {x}^{2}$ both sides:

$\rightarrow 2 {x}^{2} + 5 x - 3 = 0$

Now this is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

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

:)