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

The solutions to the equation (2x-1)^2+(1-x)^2=3 are x=(3+sqrt(14))/5 ,x=(3-sqrt(14))/5

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

The zeros of (2x-1)^2+(1-x)^2=3 are x=(3+sqrt(14))/5 ,x=(3-sqrt(14))/5

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(2 x - 1)^{2} + (1 - x)^{2} = 3

x = \frac{3 + 14}{5}, x = \frac{3 - 14}{5}

Decimal Notation

x = 1.34833 \dots, x = - 0.14833 \dots

Solution steps

(2 x - 1)^{2} + (1 - x)^{2} = 3

Expand

(2 x - 1)^{2} + (1 - x)^{2} : 5 x^{2} - 6 x + 2

5 x^{2} - 6 x + 2 = 3

Move

3

to the left side

5 x^{2} - 6 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

(- 6)^{2} - 4 \cdot 5 (- 1) = 214

x_{1, 2} = \frac{- ( - 6 ) \pm 2 14}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 14}{2 \cdot 5}, x_{2} = \frac{- ( - 6 ) - 2 14}{2 \cdot 5}

x = \frac{- ( - 6 ) + 2 14}{2 \cdot 5} : \frac{3 + 14}{5}

x = \frac{- ( - 6 ) - 2 14}{2 \cdot 5} : \frac{3 - 14}{5}

The solutions to the quadratic equation are:

x = \frac{3 + 14}{5}, x = \frac{3 - 14}{5}

i^{76}

p^{2} - 14 p

2 x^{2} - 6 x - 5 = 0

1 ln (1) - 1

- \frac{3}{2} x^{2} - 3 x - \frac{1}{2} = - \frac{3}{8} x^{2} - \frac{3}{4} x - \frac{31}{8}

What are the solutions to the equation (2x-1)^2+(1-x)^2=3 ?
The solutions to the equation (2x-1)^2+(1-x)^2=3 are x=(3+sqrt(14))/5 ,x=(3-sqrt(14))/5
Find the zeros of (2x-1)^2+(1-x)^2=3
The zeros of (2x-1)^2+(1-x)^2=3 are x=(3+sqrt(14))/5 ,x=(3-sqrt(14))/5