Question

Solve: (2x+1/x+1)^4 -6(2x-1/x+1)^2 +8=0

Answer 1

x = ± $\dfrac{1}{\sqrt{2}}$

Step-by-step explanation:

We have,

$(\dfrac{2x+1}{x+1})^4-6(\dfrac{2x+1}{x+1})^2+8=0$         ...........(1)

To find, the value of x in the given expression.

Let $(\dfrac{2x+1}{x+1})^2$ = a

Now, equation (1) becomes,

$a^2$ - 6a + 8 = 0

⇒ $a^2$ - 4a - 2a + 8 = 0

⇒ a(a - 4) - 2(a - 4) = 0

⇒ (a - 4)(a - 2) = 0

∴ $(\dfrac{2x+1}{x+1})^2$ = 4

⇒ $(\dfrac{2x+1}{x+1})^2$ = $2^2$

⇒ $\dfrac{2x+1}{x+1}$ = 2

⇒ 2x + 1 = 2x + 2

⇒ x = 0

$(\dfrac{2x+1}{x+1})^2$ = 2

⇒ 4$x^2$ + 4x + 1 = 2($x^2$ + 2x + 1)

⇒ 4$x^2$ + 4x + 1 = 2$x^2$ + 4x + 2

⇒2$x^2$ - 1 = 0

⇒ 2$x^2$ = 1

⇒ x = ±  $\dfrac{1}{\sqrt{2}}$

Thus, x = ± $\dfrac{1}{\sqrt{2}}$

Answer 2

$\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}$

x = ± 1/√2