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

The solutions to the equation x^2+(-1/4)^2=1^2 are x=(sqrt(15))/4 ,x=-(sqrt(15))/4

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

The zeros of x^2+(-1/4)^2=1^2 are x=(sqrt(15))/4 ,x=-(sqrt(15))/4

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

x = \frac{15}{4}, x = - \frac{15}{4}

Decimal Notation

x = 0.96824 \dots, x = - 0.96824 \dots

Solution steps

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

Expand

x^{2} + (- \frac{1}{4})^{2} : x^{2} + \frac{1}{16}

Expand

1^{2} : 1

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

Move

\frac{1}{16}

to the right side

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

Simplify

1 - \frac{1}{16} : \frac{15}{16}

For

x^{2} = f (a)

the solutions are

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

x = \frac{15}{16}, x = - \frac{15}{16}

\frac{15}{16} = \frac{15}{4}

- \frac{15}{16} = - \frac{15}{4}

x = \frac{15}{4}, x = - \frac{15}{4}

5 x - (x + 1) = 5 - 2 x

\frac{3 + x}{3 - x} \geq 1

\frac{c}{c ^{2} - d ^{2}} + \frac{d}{c ^{2}}

∣ m ∣ + 2 = 11

x^{2} \geq 0

What are the solutions to the equation x^2+(-1/4)^2=1^2 ?
The solutions to the equation x^2+(-1/4)^2=1^2 are x=(sqrt(15))/4 ,x=-(sqrt(15))/4
Find the zeros of x^2+(-1/4)^2=1^2
The zeros of x^2+(-1/4)^2=1^2 are x=(sqrt(15))/4 ,x=-(sqrt(15))/4