What are the solutions to the equation 4x^2-8x=1 ?

The solutions to the equation 4x^2-8x=1 are x=(2+sqrt(5))/2 ,x=(2-sqrt(5))/2

Find the zeros of 4x^2-8x=1

The zeros of 4x^2-8x=1 are x=(2+sqrt(5))/2 ,x=(2-sqrt(5))/2

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4 x^{2} - 8 x = 1

x = \frac{2 + 5}{2}, x = \frac{2 - 5}{2}

Decimal Notation

x = 2.11803 \dots, x = - 0.11803 \dots

Solution steps

4 x^{2} - 8 x = 1

Move

1

to the left side

4 x^{2} - 8 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 (- 1) = 45

x_{1, 2} = \frac{- ( - 8 ) \pm 4 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 5}{2 \cdot 4}, x_{2} = \frac{- ( - 8 ) - 4 5}{2 \cdot 4}

x = \frac{- ( - 8 ) + 4 5}{2 \cdot 4} : \frac{2 + 5}{2}

x = \frac{- ( - 8 ) - 4 5}{2 \cdot 4} : \frac{2 - 5}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 5}{2}, x = \frac{2 - 5}{2}

\frac{t + 1}{t - 1} \geq 0

∣ 8 n - 2 ∣ - 6 < 24

(\frac{3}{4} - \frac{1}{5}) \cdot \frac{3}{4}

x^{4} \leq 256 + 16 x - x^{3}

x + 2 \leq 8

What are the solutions to the equation 4x^2-8x=1 ?
The solutions to the equation 4x^2-8x=1 are x=(2+sqrt(5))/2 ,x=(2-sqrt(5))/2
Find the zeros of 4x^2-8x=1
The zeros of 4x^2-8x=1 are x=(2+sqrt(5))/2 ,x=(2-sqrt(5))/2