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प्रश्न
Find the roots of the following equation, if they exist, by applying the quadratic formula:
`1/x - 1/(x - 2) = 3, x ≠ 0, 2`
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उत्तर
The given equation is
`1/x - 1/(x - 2) = 3, x ≠ 0, 2`
⇒ `(x - 2 - x)/(x(x - 2)) = 3`
⇒ `-2/(x^2 - 2x) = 3`
⇒ –2 = 3x2 – 6x
⇒ 3x2 – 6x + 2 = 0
This equation is of the form ax2 + bx + c = 0 Where a = 3, b = –6 and c = 2.
∴ Discriminant, D = b2 – 4ac
= (–6)2 – 4 × 3 × 2
= 36 – 24
= 12 > 0
So, the given equation has real roots.
Now, `sqrt(D) = sqrt(12) = 2sqrt(3)`
∴ `α = (-b + sqrt(D))/(2a)`
= `(-(-6) + 2sqrt(3))/(2 xx 3)`
= `(6 + 2sqrt(3))/6`
= `(3 + sqrt(3))/3`
`β = (-b - sqrt(D))/(2a)`
= `(-(-6)-2sqrt(3))/(2 xx 3)`
= `(6 - 2sqrt(3))/6`
= `(3 - sqrt(3))/3`
Hence, `(3 + sqrt(3))/3` and `(3 - sqrt(3))/3` are the roots of the given equation.
