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

The solutions to the equation ((2y-3)/3)^2+y^2=1 are y= 12/13 ,y=0

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

The zeros of ((2y-3)/3)^2+y^2=1 are y= 12/13 ,y=0

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

y = \frac{12}{13}, y = 0

Decimal Notation

y = 0.92307 \dots, y = 0

Solution steps

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

Expand

(\frac{2 y - 3}{3})^{2} + y^{2} : \frac{13 y ^{2}}{9} - \frac{4 y}{3} + 1

\frac{13 y ^{2}}{9} - \frac{4 y}{3} + 1 = 1

Move

1

to the left side

\frac{13 y ^{2}}{9} - \frac{4 y}{3} = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm ( - \frac{4}{3} ) ^{2} - 4 \cdot \frac{13}{9} \cdot 0}{2 \cdot \frac{13}{9}}

(- \frac{4}{3})^{2} - 4 \cdot \frac{13}{9} \cdot 0 = \frac{4}{3}

y_{1, 2} = \frac{- ( - \frac{4}{3} ) \pm \frac{4}{3}}{2 \cdot \frac{13}{9}}

Separate the solutions

y_{1} = \frac{- ( - \frac{4}{3} ) + \frac{4}{3}}{2 \cdot \frac{13}{9}}, y_{2} = \frac{- ( - \frac{4}{3} ) - \frac{4}{3}}{2 \cdot \frac{13}{9}}

y = \frac{- ( - \frac{4}{3} ) + \frac{4}{3}}{2 \frac{13}{9}} : \frac{12}{13}

y = \frac{- ( - \frac{4}{3} ) - \frac{4}{3}}{2 \frac{13}{9}} : 0

The solutions to the quadratic equation are:

y = \frac{12}{13}, y = 0

t^{2} - 3 t - 1 = 0

5 x = x^{2} + 4

3 x (x + 1) - 2 = 16

- 5 t^{2} + 14 t + 3 = 0

4 x^{2} - 11 x + 3 = 0

What are the solutions to the equation ((2y-3)/3)^2+y^2=1 ?
The solutions to the equation ((2y-3)/3)^2+y^2=1 are y= 12/13 ,y=0
Find the zeros of ((2y-3)/3)^2+y^2=1
The zeros of ((2y-3)/3)^2+y^2=1 are y= 12/13 ,y=0