What are the solutions to the equation 1=y^2+((3y)/4)^2 ?

The solutions to the equation 1=y^2+((3y)/4)^2 are y= 4/5 ,y=-4/5

Find the zeros of 1=y^2+((3y)/4)^2

The zeros of 1=y^2+((3y)/4)^2 are y= 4/5 ,y=-4/5

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

y = \frac{4}{5}, y = - \frac{4}{5}

Decimal Notation

y = 0.8, y = - 0.8

Solution steps

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

Expand

y^{2} + (\frac{3 y}{4})^{2} : \frac{25 y ^{2}}{16}

1 = \frac{25 y ^{2}}{16}

Switch sides

\frac{25 y ^{2}}{16} = 1

Move

1

to the left side

\frac{25 y ^{2}}{16} - 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot \frac{25}{16} ( - 1 )}{2 \cdot \frac{25}{16}}

0^{2} - 4 \cdot \frac{25}{16} (- 1) = \frac{5}{2}

y_{1, 2} = \frac{- 0 \pm \frac{5}{2}}{2 \cdot \frac{25}{16}}

Separate the solutions

y_{1} = \frac{- 0 + \frac{5}{2}}{2 \cdot \frac{25}{16}}, y_{2} = \frac{- 0 - \frac{5}{2}}{2 \cdot \frac{25}{16}}

y = \frac{- 0 + \frac{5}{2}}{2 \frac{25}{16}} : \frac{4}{5}

y = \frac{- 0 - \frac{5}{2}}{2 \frac{25}{16}} : - \frac{4}{5}

The solutions to the quadratic equation are:

y = \frac{4}{5}, y = - \frac{4}{5}

x^{2} + 4 x - 9 = x + 1

4 x^{2} \cdot 4 x^{2}

8 x^{2} + 112 x + 392

x^{2} + 3 = 2 x

\frac{4}{7} x - \frac{10}{7} = - \frac{7}{4} x + \frac{1}{4}

What are the solutions to the equation 1=y^2+((3y)/4)^2 ?
The solutions to the equation 1=y^2+((3y)/4)^2 are y= 4/5 ,y=-4/5
Find the zeros of 1=y^2+((3y)/4)^2
The zeros of 1=y^2+((3y)/4)^2 are y= 4/5 ,y=-4/5