Question

Solve the following quadratic equations : $\frac{1}{x} -\frac{1}{x-2}= 3$, x≠0, 2

Answer 1

SOLUTION :  

Given : 1/x -  1/(x -2) = 3

(1 × (x-2) - x)/x(x - 2)  = 3

[By taking LCM]

(x - 2 - x )/x² - 2x = 3

-2 = 3  (x² - 2x)

3x² - 6x = -2  

3x² - 6x + 2 = 0

By using quadratic formula :  

x = [- b ±√b² - 4ac]/2a

Here, a = 3 , b= - 6 , c = 2

x = [-(-6) ± √(-6)² - 4 × 3 × 2]/2×3

x = [6 ± √36 - 24]/6

x = [ 6 ± √12]/6

x = [6 ± √4 × 3]/6

x = [6 ± 2√3]/6

x = [2(3 ± √3]/6

x = (3±√3)/3

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

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Solution :

Given ,

$\frac{1}{x} -\frac{1}{x-2}= 3$, x≠0, 2

=> [x-2-x]/[x(x-2)] = 3

=> -2/(x²-2x) = 3

=> -2 = 3(x²-2x)

=> -2 = 3x²-6x

=> 3x² - 6x + 2 = 0

Compare above equation

with ax² + bx + c = 0 , we get

a = 3 , b = -6 , c = 2

Discreminant (D) = b²-4ac

= (-6)² - 4×3×2

= 36 - 24

= 12

Now ,

By Quadratic Formula :

x = [-b±√D]/2a

= [-(-6)±√12]/(2×3)

= ( 6 ± 2√3 )/6

= [2(3±√3)/6

= ( 3 ± √3 )/3

Therefore ,

x = ( 3 + √3 )/3 Or

x = ( 3-√3 )/3

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