Question

$x + \frac{1}{x} = 3 \times \frac{1}{3}$
where x is not equal to 0​

Answer 1

$\huge\underline\mathcal{Question}$

${x+}\frac{1}{x}=3×\frac{1}{3}$

$\huge\underline\mathcal{To\:find}$

Find the value of x

$\huge\underline\mathcal{Solution}$

$\implies\sf x+\frac{1}{x}=3×\frac{1}{3}$

$\implies\sf \frac{x^2+1}{x}=1$

$\implies\sf x^2+1=x$

$\implies\sf x^2-x+1=0$

Applying quadratic formula to solve this equation

As we know that

D = b² - 4ac

a = 1 b = -1 c = 1

putting these value in above formula

D = b² - 4ac

D = (-1)²- 4×1×1

D = 1-4

D = -3

This equation have no real roots