Question

$x \: + {x}^{ - 1} = \: 3 \\ \\ find \: the \: value \: of \: x$
Solve for x


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Give useful and appropriate ans

Answer 1

Hey !!!

here is your answer
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$x \: + {x}^{ - 1} = \: 3 \\ \\ x \: + \frac{1}{x} - \: 3 = 0 \\ \\ \frac{ {x}^{2} + 1 + 3}{x} = 0 \\ \\ {x}^{2} - 3x + 1 = 0 \\ \\ this \: is \: in \: the \: form \: of \: quardtaic \: \\ eqution \: in \: which \: a \: = 1 \: \\ b \: = - 3 \\ c = 1 \\ \\ using \: quardic \: formula \\ \\ \\ d = \frac{ - b \: + - \sqrt{ } { b}^{2} - 4ac }{2a }$

［ - (-3) ± √( - 3)² - 4 . 1 . 1 ］/ 2.1

［ 3 ± √ 9 - 4 ］/ 2

either ,

［ 3 + √ 5 ］/ 2

........or

［ 3 - √5 ］/ 2

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Answer 2

Answer :

For a quadratic equation ax² + bx + c = 0, the roots be

x = {- b ± √(b² - 4ac) }/(2a)

The given equation is

x + x⁻¹ = 3

⇒ x + 1/x = 3

⇒ x² - 3x + 1 = 0

Using the above formula, we get

x = [ - (- 3) ± √{(- 3)² - (4 × 1 × 1)} ]/(2 × 1)

= {3 ± √(9 - 4) }/2

= (3 ± √5)/2

∴ the required roots of the given equation are

x = (3 + √5)/2, (3 - √5)/2

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