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

The solutions to the equation x^2+(3/4)^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4

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

The zeros of x^2+(3/4)^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4

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

x = \frac{7}{4}, x = - \frac{7}{4}

Decimal Notation

x = 0.66143 \dots, x = - 0.66143 \dots

Solution steps

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

Expand

x^{2} + (\frac{3}{4})^{2} : x^{2} + \frac{9}{16}

x^{2} + \frac{9}{16} = 1

Move

\frac{9}{16}

to the right side

x^{2} = 1 - \frac{9}{16}

Simplify

1 - \frac{9}{16} : \frac{7}{16}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{7}{16}, x = - \frac{7}{16}

\frac{7}{16} = \frac{7}{4}

- \frac{7}{16} = - \frac{7}{4}

x = \frac{7}{4}, x = - \frac{7}{4}

x^{2} - 2 x - 8

x (x - 1) = 1

x^{2} - x - 12 = 0

4 x^{2} + 5 b x = - b^{2}

2 x - 3 x^{2} = 5

What are the solutions to the equation x^2+(3/4)^2=1 ?
The solutions to the equation x^2+(3/4)^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4
Find the zeros of x^2+(3/4)^2=1
The zeros of x^2+(3/4)^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4