Question

x+1/x =14 then x√x-14 = ?​

Answer 1

Answer:

-2√2

Step-by-step explanation:

m-1/m ko solve karke m ki value place kar

Answer 2

Given:

$x+\dfrac{1}{x}$ = 14

We have to find, $\sqrt{x} +\dfrac{1}{\sqrt{x} } -4$ = ?

Solution:

$x+\dfrac{1}{x}$ = 14

⇒ $(\sqrt{x})^2 +(\dfrac{1}{\sqrt{x} } )^2$ = 14

Using the algebraic identity:

$(a+b)^{2}$= $a^{2} +b^{2}$ + 2ab

⇒ $a^{2} +b^{2}$ = $(a+b)^{2}$ - 2ab

$(\sqrt{x} +\dfrac{1}{\sqrt{x} })^2-2(\sqrt{x} )(\dfrac{1}{\sqrt{x} })$ = 14

⇒ $(\sqrt{x} +\dfrac{1}{\sqrt{x} })^2$- 2 = 14

⇒ $(\sqrt{x} +\dfrac{1}{\sqrt{x} })^2$ = 14 + 2

⇒ $(\sqrt{x} +\dfrac{1}{\sqrt{x} })^2$ = 16

⇒ $(\sqrt{x} +\dfrac{1}{\sqrt{x} })^2$ = $4^2$

⇒ $\sqrt{x} +\dfrac{1}{\sqrt{x} }$ = 4

Subtracting both sides by 4, we get

$\sqrt{x} +\dfrac{1}{\sqrt{x} }$  - 4 = 4 - 4

⇒ $\sqrt{x} +\dfrac{1}{\sqrt{x} }$  - 4 = 0

∴ $\sqrt{x} +\dfrac{1}{\sqrt{x} }$  - 4 = 0

Thus, if $x+\dfrac{1}{x}$ = 14, then $\sqrt{x} +\dfrac{1}{\sqrt{x} }$  - 4 = 0.