Question

How do you solve `x^2=25/289`?

Answer 1

`x=+-(5/17) approx +-0.294116`

Explanation:

`x^2 = 25/289`

`x=+-sqrt(25/289)= +- sqrt25/sqrt289`

Note: `25 = 5^2 and 289=17^2`

`:. x=+-5/17 approx +-0.294116`