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If x2 - 16x + 59 = 0, then what is the value of (x - 6)2 + [1/(x - 6)2]?
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- B18
- C16
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Discussion 1
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CCommunity
Let (x - 6) = t
⇒ x = t + 6 ----(1)
x2 - 16x + 59 = 0
Putting value of x from equation (1)
⇒ (t + 6)2 - 16(t + 6) + 59 = 0
⇒ t2 + 36 + 12t - 16t - 96 + 59 = 0
⇒ t2 - 4t - 1 = 0
On dividing the above equation by t,
⇒ t - 4 - 1/t = 0
⇒ t - 1/t = 4
On squaring above equation,
⇒ (t - 1/t)2 = 42
⇒ t2 + 1/t2 - 2 = 16
⇒ t2 + 1/t2 = 18
∴ (x - 6)2 + [1/(x - 6)2] = 18
⇒ x = t + 6 ----(1)
x2 - 16x + 59 = 0
Putting value of x from equation (1)
⇒ (t + 6)2 - 16(t + 6) + 59 = 0
⇒ t2 + 36 + 12t - 16t - 96 + 59 = 0
⇒ t2 - 4t - 1 = 0
On dividing the above equation by t,
⇒ t - 4 - 1/t = 0
⇒ t - 1/t = 4
On squaring above equation,
⇒ (t - 1/t)2 = 42
⇒ t2 + 1/t2 - 2 = 16
⇒ t2 + 1/t2 = 18
∴ (x - 6)2 + [1/(x - 6)2] = 18