if \({P^2} + {1 \over {{P^2}}} = 7\) , then find the value of \({P^2} - {1 \over {{P^2}}}.\)
Solution
Correct Answer: Option B
\(\eqalign{ & {\left( {a + b} \right)^2} = {\left( {a - b} \right)^2} + 4ab \cr & \Rightarrow\ \ \ \ \ {\left( {P + {1 \over P}} \right)^2} = {\left( {P - {1 \over P}} \right)^2} + 4 \cr}\)
Given expression:
\(\Rightarrow\ \ \ \ {P^2} + {1 \over {{P^2}}} = 7\) -------------- (1)
Subtracting 2 from both sides, we get:
\(\eqalign{&\Rightarrow\ \ \ \ {P^2} + {1 \over {{P^2}}} - 2 = 7 - 2 \cr &\Rightarrow\ \ \ \ \ {\left( {P - {1 \over P}} \right)^2} = 5 \cr}\)
\(\Rightarrow\ \ \ \ \ P - {1 \over P} = \sqrt 5\) -------------- (2)
Now adding 2 to both sides in equation (1) we get,
\(\eqalign{ & \Rightarrow\ \ \ \ \ {P^2} + {1 \over {{P^2}}} + 2 = 7 + 2 \cr & \Rightarrow\ \ \ \ {\left( {P + {1 \over P}} \right)^2} = 9 \cr}\)
\(\Rightarrow P + {1 \over P} = 3\) -------------- (3)
Formula: -
\(\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)\)
Now, multiplying equations (2) and (3) we get,
\(\eqalign{ & \Rightarrow\ \ \ \ \left( {P + {1 \over P}} \right)\left( {P - {1 \over P}} \right) = 3\sqrt 5 \cr & \Rightarrow\ \ \ \ {P^2} - {1 \over {{P^2}}} = 3\sqrt 5 \cr}\)
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