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Question
Solve the following by reducing them to quadratic equations:
`((7y - 1)/y)^2 - 3 ((7y - 1)/y) - 18 = 0, y ≠ 0`
Solution
The given equation
`((7y - 1)/y)^2 - 3 ((7y - 1)/y) - 18 = 0, y ≠ 0`
Putting `(7y - 1)/y`` = z`, then given equation becomes
z2 - 3z - 18 = 0
⇒ z2 - 6z + 3z - 18 = 0
⇒ z(z - 6) + 3(z - 6) = 0
⇒ (z - 6) (z + 3) = 0
⇒ z - 6 = 0 or z + 3 = 0
⇒ z = 6 or z = -3
But `(7y - 1)/y =z`
∴ `(7y - 1)/y` = 6
⇒ 7y - 1
= 6y
⇒ 7y - 6y
= 1
⇒ y = 1
Also `(7y - 1)/y`
= -3
⇒ 7y - 1
= -3y
⇒ 7y + 3y -1 = 0
⇒ 10y = 1
⇒ y = `(1)/(10)`
Hence, the required roots are `(1)/(10),1`.
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