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

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

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

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

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

x = \frac{15}{2 2}, x = - \frac{15}{2 2}

Decimal Notation

x = 1.36930 \dots, x = - 1.36930 \dots

Solution steps

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

Move

4

to the right side

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

Move

x^{2}

to the left side

- 2 x^{2} = \frac{1}{4} - 4

Simplify

\frac{1}{4} - 4 : - \frac{15}{4}

- 2 x^{2} = - \frac{15}{4}

Divide both sides by

- 2

x^{2} = \frac{15}{8}

For

x^{2} = f (a)

the solutions are

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

x = \frac{15}{8}, x = - \frac{15}{8}

\frac{15}{8} = \frac{15}{2 2}

- \frac{15}{8} = - \frac{15}{2 2}

x = \frac{15}{2 2}, x = - \frac{15}{2 2}

- 4 = x^{2} - 4 x

6 x^{2} + 8 x

x^{2} + x - \frac{1}{4} = 0

r^{2} + r + 6 = 0

y, (x - 3)^{2} + (y + 2)^{2} = 4

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