What are the solutions to the equation solvefor P,P^2-P(1/l+2l)+2=0 ?

The solutions to the equation solvefor P,P^2-P(1/l+2l)+2=0 are P=2l,P= 1/l

Find the zeros of solvefor P,P^2-P(1/l+2l)+2=0

The zeros of solvefor P,P^2-P(1/l+2l)+2=0 are P=2l,P= 1/l

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P, P^{2} - P (\frac{1}{l} + 2 l) + 2 = 0

P = 2 l, P = \frac{1}{l}

Solution steps

P^{2} - P (\frac{1}{l} + 2 l) + 2 = 0

Write in the standard form

a x^{2} + b x + c = 0

P^{2} - (\frac{1}{l} + 2 l) P + 2 = 0

Solve with the quadratic formula

P_{1, 2} = \frac{- ( - \frac{1}{l} - 2 l ) \pm ( - \frac{1}{l} - 2 l ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

Simplify

(- \frac{1}{l} - 2 l)^{2} - 4 \cdot 1 \cdot 2 : \frac{( 2 l + 1 ) ( 2 l - 1 )}{l}

P_{1, 2} = \frac{- ( - \frac{1}{l} - 2 l ) \pm \frac{( 2 l + 1 ) ( 2 l - 1 )}{l}}{2 \cdot 1}

Separate the solutions

P_{1} = \frac{- ( - \frac{1}{l} - 2 l ) + \frac{( 2 l + 1 ) ( 2 l - 1 )}{l}}{2 \cdot 1}, P_{2} = \frac{- ( - \frac{1}{l} - 2 l ) - \frac{( 2 l + 1 ) ( 2 l - 1 )}{l}}{2 \cdot 1}

P = \frac{- ( - \frac{1}{l} - 2 l ) + \frac{( 2 l + 1 ) ( 2 l - 1 )}{l}}{2 \cdot 1} : 2 l

P = \frac{- ( - \frac{1}{l} - 2 l ) - \frac{( 2 l + 1 ) ( 2 l - 1 )}{l}}{2 \cdot 1} : \frac{1}{l}

The solutions to the quadratic equation are:

P = 2 l, P = \frac{1}{l}

6 x^{2} + 480 = 0

x^{2} - 15 x + 12 = 0

x, x^{2} + 14 x + 49 = 0

n^{2} + 8 n = - 15

80 = 14 x^{2}

What are the solutions to the equation solvefor P,P^2-P(1/l+2l)+2=0 ?
The solutions to the equation solvefor P,P^2-P(1/l+2l)+2=0 are P=2l,P= 1/l
Find the zeros of solvefor P,P^2-P(1/l+2l)+2=0
The zeros of solvefor P,P^2-P(1/l+2l)+2=0 are P=2l,P= 1/l