What are the solutions to the equation 4x^2-x+1=x+3x^2 ?

The solutions to the equation 4x^2-x+1=x+3x^2 are x=1

Find the zeros of 4x^2-x+1=x+3x^2

The zeros of 4x^2-x+1=x+3x^2 are x=1

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4 x^{2} - x + 1 = x + 3 x^{2}

x = 1

Solution steps

4 x^{2} - x + 1 = x + 3 x^{2}

Move

3 x^{2}

to the left side

x^{2} - x + 1 = x

Move

x

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

x = \frac{- ( - 2 )}{2 \cdot 1}

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

y^{2} = - 5

x, x^{2} = 9

36 x^{2} + 4 = 0

\frac{7 x ^{2} - 5 x}{3} = \frac{x ^{2} + 4 x}{6}

r^{2} + 18 r + 56 = 0

What are the solutions to the equation 4x^2-x+1=x+3x^2 ?
The solutions to the equation 4x^2-x+1=x+3x^2 are x=1
Find the zeros of 4x^2-x+1=x+3x^2
The zeros of 4x^2-x+1=x+3x^2 are x=1