Asquare is inscribed in a right isosceles triangle, such that two of its vertices lie on the hypotenuse and two other on the legs. find the length of the side of the square, if the length of the hypotenuse is 3 in.

The explanation of the problem is shown in the attached figure.
Let the side length of the square be x.
The triangular regions outside the square are all isosceles right triangles
The length of the hypotenuse is 3 in
So,
     The length of each leg = 3/√2
     The area of the given triangle = (1/2) * (3/√2)*(3/√2) = 9/4
With the help of figure:

Area1= (1/2) * (x/√2)*(x/√2) = x²/4
Area2= (1/2) * (x)*(x) = x²/2
Area3= (1/2) * (x)*(x) = x²/2
Area of the square = x²

So, Area1 + Area2 + Area3 + square area = 9/4
∴
    x²/4 + x²/2 + x²/2 +x² = 9/4
   (9/4) x² = 9/4
   x² = 1
∴ x = 1

∴ The side of the square which is inscribed in a right isosceles triangle = 1 in

X= -5because this is a single step equation so you would just have to divide -16 both sides for x to equal your solution