What are the solutions to the equation x-x^2=log_{5}(1/25) ?

The solutions to the equation x-x^2=log_{5}(1/25) are x=-1,x=2

Find the zeros of x-x^2=log_{5}(1/25)

The zeros of x-x^2=log_{5}(1/25) are x=-1,x=2

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x - x^{2} = lo g_{5} (\frac{1}{25})

x = - 1, x = 2

Solution steps

x - x^{2} = lo g_{5} (\frac{1}{25})

Expand

lo g_{5} (\frac{1}{25}) : - 2

x - x^{2} = - 2

Move

2

to the left side

x - x^{2} + 2 = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

1^{2} - 4 (- 1) \cdot 2 = 3

x_{1, 2} = \frac{- 1 \pm 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 3}{2 ( - 1 )}, x_{2} = \frac{- 1 - 3}{2 ( - 1 )}

x = \frac{- 1 + 3}{2 ( - 1 )} : - 1

x = \frac{- 1 - 3}{2 ( - 1 )} : 2

The solutions to the quadratic equation are:

x = - 1, x = 2

7 x^{2} - 14 x + 5 = - x^{2}

a^{2} + 5 a - 14

\frac{4 x ^{- 2}}{y ^{- 8}}

12 x - 4 (5 x - 1) = 3 - 4 x

- 0.375 c - 11 = 13

What are the solutions to the equation x-x^2=log_{5}(1/25) ?
The solutions to the equation x-x^2=log_{5}(1/25) are x=-1,x=2
Find the zeros of x-x^2=log_{5}(1/25)
The zeros of x-x^2=log_{5}(1/25) are x=-1,x=2