What are the solutions to the equation x^2+16/25 x^2=1 ?

The solutions to the equation x^2+16/25 x^2=1 are x= 5/(sqrt(41)),x=-5/(sqrt(41))

Find the zeros of x^2+16/25 x^2=1

The zeros of x^2+16/25 x^2=1 are x= 5/(sqrt(41)),x=-5/(sqrt(41))

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x^{2} + \frac{16}{25} x^{2} = 1

x = \frac{5}{41}, x = - \frac{5}{41}

Decimal Notation

x = 0.78086 \dots, x = - 0.78086 \dots

Solution steps

x^{2} + \frac{16}{25} x^{2} = 1

Multiply both sides by

25

25 x^{2} + 16 x^{2} = 25

41 x^{2} = 25

Divide both sides by

41

x^{2} = \frac{25}{41}

For

x^{2} = f (a)

the solutions are

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

x = \frac{25}{41}, x = - \frac{25}{41}

\frac{25}{41} = \frac{5}{41}

- \frac{25}{41} = - \frac{5}{41}

x = \frac{5}{41}, x = - \frac{5}{41}

3 (x + 8) = x + 38

\frac{y ^{4}}{y ^{2}}

\frac{1}{\frac{- 2}{2}}

3 c^{2} - c

3^{11}

What are the solutions to the equation x^2+16/25 x^2=1 ?
The solutions to the equation x^2+16/25 x^2=1 are x= 5/(sqrt(41)),x=-5/(sqrt(41))
Find the zeros of x^2+16/25 x^2=1
The zeros of x^2+16/25 x^2=1 are x= 5/(sqrt(41)),x=-5/(sqrt(41))