What are the solutions to the equation 4a^2-4a=1 ?

The solutions to the equation 4a^2-4a=1 are a=(1+sqrt(2))/2 ,a=(1-sqrt(2))/2

Find the zeros of 4a^2-4a=1

The zeros of 4a^2-4a=1 are a=(1+sqrt(2))/2 ,a=(1-sqrt(2))/2

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4 a^{2} - 4 a = 1

a = \frac{1 + 2}{2}, a = \frac{1 - 2}{2}

Solution steps

4 a^{2} - 4 a = 1

Move

1

to the left side

4 a^{2} - 4 a - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 4 (- 1) = 42

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

Separate the solutions

a_{1} = \frac{- ( - 4 ) + 4 2}{2 \cdot 4}, a_{2} = \frac{- ( - 4 ) - 4 2}{2 \cdot 4}

a = \frac{- ( - 4 ) + 4 2}{2 \cdot 4} : \frac{1 + 2}{2}

a = \frac{- ( - 4 ) - 4 2}{2 \cdot 4} : \frac{1 - 2}{2}

The solutions to the quadratic equation are:

a = \frac{1 + 2}{2}, a = \frac{1 - 2}{2}

4 (y + 5) + 3 = 35

(\frac{x ^{3}}{y ^{2}})^{- 1}

(x^{2})^{- 1}

78. 2^{2} + 41. 3^{2} = c^{2}

\frac{x y}{30} - \frac{x y}{12}

What are the solutions to the equation 4a^2-4a=1 ?
The solutions to the equation 4a^2-4a=1 are a=(1+sqrt(2))/2 ,a=(1-sqrt(2))/2
Find the zeros of 4a^2-4a=1
The zeros of 4a^2-4a=1 are a=(1+sqrt(2))/2 ,a=(1-sqrt(2))/2