Subject

    Quantitative Aptitude

    Topic

    Algebra

    if P2+1P2=7{P^2} + {1 \over {{P^2}}} = 7 , then find the value of P2−1P2.{P^2} - {1 \over {{P^2}}}.

    ক)
    \(2\sqrt 5\)
    খ)
    \(3\sqrt 5\)
    গ)
    \(4\sqrt 5\)
    ঘ)
    \(5\sqrt 5\)
    ঙ)
    None of these

    Explanation

    \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: ⇒    P2+1P2=7\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} ⇒     P−1P=5\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} ⇒P+1P=3\Rightarrow P + {1 \over P} = 3    -------------- (3) Formula: - (a+b)(a−b)=(a2−b2)\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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