Question

Solve the following quadratic equations : $x-\frac{1}{x}=3$, x≠0

Answer 1

SOLUTION :  

Given : x - 1/x = 3  

(x² - 1)/x = 3

[By taking LCM]

x² - 1 = 3x  

x² - 3x - 1 = 0

By using quadratic formula :  

x = [- b ±√b² - 4ac]/2a

Here, a = 1 , b= - 3 , c = - 1

x = [ -(-3) ±√(-3)² - 4 × 1× -1]/2×1

x = [3 ± √9 + 4]/2

x = [3 ± √13]/2

Hence, the roots of the above quadratic equation are x = (3 + √13)/2 and x = (3 -  √13)/2

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Answer 2

Solution :

x - 1/x = 3

=> ( x² - 1 )/x = 3

=> x² - 1 = 3x

=> x² - 3x - 1 = 0

Compare this equation with

ax² + bx + c = 0 we get

a = 1 , b = -3 , c = -1

Discreminant (D) = b²-4ac

= (-3)² - 4×1×(-1)

= 9 + 4

= 13

By Quadratic Formula :

x = [-b±√D]/2a

= [-(-3)±√13]/(2×1)

= (3±√13)/2

Therefore ,

x = (3+√13)/2 or

x = (3-√13)/2

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