Question

give me correct ans with easiest solution​

Answer 1

Step-by-step explanation:

Taking x² = y

y + 1/y = 82/9

y²+1 = 82y/9

9y² - 82y +9 = 0

9y² - 81y - y +9 = 0

9y ( y-9) - 1 (y-9) = 0

(y-9)(9y-1) = 0

So, y = 9 or 1/9

and , x² = √9 or √1/9

x = 3 or 1/3.

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Given,

$\large{ \sf{x {}^{2} + \frac{1}{x {}^{2} } = \frac{82}{9} }}$

$\sf{Adding \ 2x. \frac{1}{x} \ on \ both \ sides}$

$\large{ \leadsto \: \sf{x {}^{2} + \frac{1}{x {}^{2} } + 2.x. \frac{1}{x} = \frac{82}{9} + 2.x. \frac{1}{x} }} \\ \\ \large{\leadsto \: \sf{x {}^{2} + \frac{1}{x {}^{2} } + 2.x. \frac{1}{x} = \frac{82}{9} + 2 }} \\ \\ \large{ \leadsto \: \sf{(x + \frac{1}{x} ) {}^{2} = \frac{100}{9} }} \\ \\ \large{ \leadsto \: \sf{x + \frac{1}{x} = \pm \sqrt{ \frac{100}{3} } }} \\ \\ \huge{ \leadsto \: \mathtt{x + \frac{1}{ x} } = \pm \: \frac{10}{3} }$

When x + 1/x = 10/3,

$\implies \: \sf{ x + \: \frac{1}{x} = \frac{10}{3} } \\ \\ \implies \: \sf{3x {}^{2} - 10x + 3 = 0 } \\ \\ \implies \: \sf{3x {}^{2} - 9x - x + 3 = 0} \\ \\ \implies \: \sf{(x - 3)(3x - 1) = 0} \\ \\ \huge{ \implies \: \underline{ \boxed{\sf{x = 3 \: or \: \frac{1}{3} }}}}$

When x + 1/x = -10/3,

$\implies \: \sf{x + \frac{1}{x} = - \frac{ 10}{3} } \\ \\ \implies \: \sf{3x {}^{2} + 3 = - 10x } \\ \\ \implies \: \sf{3x {}^{2} + 10x + 3 = 0}$

On comparing with ax² + bx + c = 0,

a = c = 3 and b = 10

Discriminant,d :

$\sf{d = (10) {}^{2} - 4(3)(3)} \\ \\ \sf{d = 100 - 36} \\ \\ \huge{ \boxed{ \boxed{ \sf{d = 64}}}}$

Applying the Quadratic Formula,

$\sf{x = \frac{ - b \pm \sqrt{d} }{2a} } \\ \\ \rightarrow \: \sf{x = \frac{ - 10 \pm \: \sqrt{64} }{6} } \\ \\ \rightarrow \: \sf{x = \frac{ - 10 \pm8}{6} } \\ \\ \rightarrow \: \sf{x = \frac{ - 5 \pm \: 4}{3} } \\ \\ \rightarrow \ \sf{x = - \frac{9}{3} \: \: or \: \: - \frac{ 1}{3} } \\ \\ \huge{ \rightarrow \: \underline{ \boxed{\sf{x = - 3 \: or \: - \frac{ 1}{3} }}}}$