Question

2(x/x+1)^2-5(x/x+1)+2=0,x is not equal to -1​

Answer 1

Step-by-step explanation:

let x/x+1 =y

2y²-5y+2 =0

2y²-4y-y+2 =0

2y(y-2) -1(y-2) =0

(2y-1)(y-2)=0

y=1/2, 2

x/(x+1) = 2

x=2x+2

x=-2

x/x+1 = 1/2

2x = x+1

x = 1

Answer 2

Given:-

• $\sf 2 {\bigg( \dfrac{x}{x + 1} \bigg)}^{2} - 5\bigg( \dfrac{x}{x + 1} \bigg) + 2 = 0$
• $\sf x \ne - 1$

To find:-

• The values of x.

Answer:-

Given that,

$\sf 2 {\bigg( \dfrac{x}{x + 1} \bigg)}^{2} - 5\bigg( \dfrac{x}{x + 1} \bigg) + 2 = 0$

Let $\sf \dfrac{x}{x + 1}$ be $\sf a.$

$\sf \longrightarrow 2 {a}^{2} - 5a + 2 = 0$

On splitting the middle term,

$\sf \longrightarrow 2 {a}^{2} - 4a - a + 2 = 0$

$\sf \longrightarrow 2a(a - 2) -( a - 2) = 0$

$\sf \longrightarrow (a - 2)(2a - 1) = 0$

So,

If (a - 2) = 0,

$\sf \: a = 2$

Substituting the value of $\sf a$ which we assumed earlier,

$\sf \longrightarrow \dfrac{x}{x + 1} = 2$

$\sf \longrightarrow x = 2(x + 1)$

$\sf \longrightarrow x = 2x + 2$

$\longrightarrow \underline{ \underline{ \sf { \green{ x_{1} = - 2}}}}$

If (2a -1) = 0,

$\sf \: a = \dfrac{1}{2}$

Substituting the value of $\sf a$ which we assumed earlier,

$\sf \longrightarrow \dfrac{x}{x + 1} = \dfrac{1}{2}$

$\sf \longrightarrow 2x = x + 1$

$\longrightarrow \underline{ \underline{ \sf { \green{ x_{2} = 1}}}}$