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

The solutions to the equation 4x-1=x^2 are x=2+sqrt(3),x=2-sqrt(3)

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

The zeros of 4x-1=x^2 are x=2+sqrt(3),x=2-sqrt(3)

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4 x - 1 = x^{2}

x = 2 + 3, x = 2 - 3

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

4 x - 1 = x^{2}

Switch sides

x^{2} = 4 x - 1

Move

1

to the left side

x^{2} + 1 = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

x^{2} + 2 x = 0

2 x^{2} - 16 x + 38

m, 6 m^{2} - 8 m + 1 = 0

6 x^{2} - 5 x - 2 = - 6 x^{2}

4 (x + 7)^{2} = 13

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