What are the solutions to the equation ((1+3y)/2)^2+2y^2=8 ?

The solutions to the equation ((1+3y)/2)^2+2y^2=8 are y=(-3+2sqrt(134))/(17),y=(-3-2sqrt(134))/(17)

Find the zeros of ((1+3y)/2)^2+2y^2=8

The zeros of ((1+3y)/2)^2+2y^2=8 are y=(-3+2sqrt(134))/(17),y=(-3-2sqrt(134))/(17)

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(\frac{1 + 3 y}{2})^{2} + 2 y^{2} = 8

y = \frac{- 3 + 2 134}{17}, y = \frac{- 3 - 2 134}{17}

Decimal Notation

y = 1.18539 \dots, y = - 1.53833 \dots

Solution steps

(\frac{1 + 3 y}{2})^{2} + 2 y^{2} = 8

Expand

(\frac{1 + 3 y}{2})^{2} + 2 y^{2} : \frac{1}{4} + \frac{3 y}{2} + \frac{17 y ^{2}}{4}

\frac{1}{4} + \frac{3 y}{2} + \frac{17 y ^{2}}{4} = 8

Move

8

to the left side

\frac{1 + 17 y ^{2}}{4} + \frac{3 y}{2} - 8 = 0

Write in the standard form

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

\frac{17 y ^{2}}{4} + \frac{3 y}{2} + \frac{1}{4} - 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- \frac{3}{2} \pm ( \frac{3}{2} ) ^{2} - 4 \cdot \frac{17}{4} ( \frac{1}{4} - 8 )}{2 \cdot \frac{17}{4}}

(\frac{3}{2})^{2} - 4 \cdot \frac{17}{4} (\frac{1}{4} - 8) = 134

y_{1, 2} = \frac{- \frac{3}{2} \pm 134}{2 \cdot \frac{17}{4}}

Separate the solutions

y_{1} = \frac{- \frac{3}{2} + 134}{2 \cdot \frac{17}{4}}, y_{2} = \frac{- \frac{3}{2} - 134}{2 \cdot \frac{17}{4}}

y = \frac{- \frac{3}{2} + 134}{2 \frac{17}{4}} : \frac{- 3 + 2 134}{17}

y = \frac{- \frac{3}{2} - 134}{2 \frac{17}{4}} : \frac{- 3 - 2 134}{17}

The solutions to the quadratic equation are:

y = \frac{- 3 + 2 134}{17}, y = \frac{- 3 - 2 134}{17}

10 x^{2} + 11 x + 3 = 0

x, (z - 4)^{2} = 4 x^{2}

2 x^{2} + 7 x - 15 = 0

2 (x^{2} - 1) = x^{2} + 7

2 = x - 2 (x + 4)^{2}

What are the solutions to the equation ((1+3y)/2)^2+2y^2=8 ?
The solutions to the equation ((1+3y)/2)^2+2y^2=8 are y=(-3+2sqrt(134))/(17),y=(-3-2sqrt(134))/(17)
Find the zeros of ((1+3y)/2)^2+2y^2=8
The zeros of ((1+3y)/2)^2+2y^2=8 are y=(-3+2sqrt(134))/(17),y=(-3-2sqrt(134))/(17)