What are the solutions to the equation 5x=2x^2+1 ?

The solutions to the equation 5x=2x^2+1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

Find the zeros of 5x=2x^2+1

The zeros of 5x=2x^2+1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

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5 x = 2 x^{2} + 1

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

Solution steps

5 x = 2 x^{2} + 1

Switch sides

2 x^{2} + 1 = 5 x

Move

5 x

to the left side

2 x^{2} + 1 - 5 x = 0

Write in the standard form

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

2 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

x_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

x = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

x = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

4 b^{2} - 19 b + 17 = 0

x^{3} - 4 x^{4}

∣ 2 x - 8 ∣ > 6

\frac{2 x - 1}{1 - x} > 3

What are the solutions to the equation 5x=2x^2+1 ?
The solutions to the equation 5x=2x^2+1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4
Find the zeros of 5x=2x^2+1
The zeros of 5x=2x^2+1 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4