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

The solutions to the equation x^2-x+1/4 = x/4 are x=1,x= 1/4

Find the zeros of x^2-x+1/4 = x/4

The zeros of x^2-x+1/4 = x/4 are x=1,x= 1/4

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x^{2} - x + \frac{1}{4} = \frac{x}{4}

x = 1, x = \frac{1}{4}

Decimal Notation

x = 1, x = 0.25

Solution steps

x^{2} - x + \frac{1}{4} = \frac{x}{4}

Multiply both sides by

4

4 x^{2} - 4 x + 1 = x

Move

x

to the left side

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 4 \cdot 1 = 3

x_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 4}

x = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

x = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

The solutions to the quadratic equation are:

x = 1, x = \frac{1}{4}

9 x^{2} + 6 x - 1 = 0

x^{2} - 22 x

2 y^{2} + 19 y + 24 = 0

- (x + 9)^{2} = 64

2 x - x^{2} + 3

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