Question

What’s X2 25/289 ? For real number in math

Answer 1

we are given

$x^2= \frac{25}{289}  $

Firstly, we will find all possible factors of 25 and 289

25=5*5

289=17*17

so, we get

$x^2= \frac{5*5}{17*17}  $

$x^2= \frac{5^2}{17^2}$

Since, we have to solve for x

so, we can take sqrt both sides

$\sqrt{x^2}= -\sqrt{\frac{5^2}{17^2} }  $

$\sqrt{x^2}= \sqrt{\frac{5^2}{17^2} }  $

now, we can simplify both

$x=-\frac{5}{17}$

$x=\frac{5}{17}$

so, solutions are

$x=-\frac{5}{17}$

$x=\frac{5}{17}$................Answer