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

The solutions to the equation v^2+(-2/3+v)^2=1 are v=(2+sqrt(14))/6 ,v=(2-sqrt(14))/6

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

The zeros of v^2+(-2/3+v)^2=1 are v=(2+sqrt(14))/6 ,v=(2-sqrt(14))/6

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v^{2} + (- \frac{2}{3} + v)^{2} = 1

v = \frac{2 + 14}{6}, v = \frac{2 - 14}{6}

Solution steps

v^{2} + (- \frac{2}{3} + v)^{2} = 1

Expand

v^{2} + (- \frac{2}{3} + v)^{2} : 2 v^{2} - \frac{4}{3} v + \frac{4}{9}

2 v^{2} - \frac{4}{3} v + \frac{4}{9} = 1

Move

1

to the left side

- \frac{5}{9} - \frac{4}{3} v + 2 v^{2} = 0

Write in the standard form

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

2 v^{2} - \frac{4 v}{3} - \frac{5}{9} = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm ( - \frac{4}{3} ) ^{2} - 4 \cdot 2 ( - \frac{5}{9} )}{2 \cdot 2}

(- \frac{4}{3})^{2} - 4 \cdot 2 (- \frac{5}{9}) = \frac{2 14}{3}

v_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm \frac{2 14}{3}}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- ( - \frac{4}{3} ) + \frac{2 14}{3}}{2 \cdot 2}, v_{2} = \frac{- ( - \frac{4}{3} ) - \frac{2 14}{3}}{2 \cdot 2}

v = \frac{- ( - \frac{4}{3} ) + \frac{2 14}{3}}{2 \cdot 2} : \frac{2 + 14}{6}

v = \frac{- ( - \frac{4}{3} ) - \frac{2 14}{3}}{2 \cdot 2} : \frac{2 - 14}{6}

The solutions to the quadratic equation are:

v = \frac{2 + 14}{6}, v = \frac{2 - 14}{6}

15 x - 2 = 85 - 4 (x - 2)

\frac{2}{7}

1 < - 6 x - 17

(x + 10) + (130) = 90

\frac{24}{8}

What are the solutions to the equation v^2+(-2/3+v)^2=1 ?
The solutions to the equation v^2+(-2/3+v)^2=1 are v=(2+sqrt(14))/6 ,v=(2-sqrt(14))/6
Find the zeros of v^2+(-2/3+v)^2=1
The zeros of v^2+(-2/3+v)^2=1 are v=(2+sqrt(14))/6 ,v=(2-sqrt(14))/6