The hypotenuse of a right triangle is $5 m$ if the smaller is doubles and longer is triples the new hypotenuse is $6\sqrt{5} m$. FInd the sides of the triangle.
What I found so far: After coming up the equation in both we will come upon a case where we will find a quadratic equation in one variable where we will have to apply Discriminant Formula or Factorisation formula to come up with the solution.
1. $$x^2 + y^2=5^2$$ $$x^2 = 25-y^2$$ 2. $$(2x)^2 + (3y)^2 = (6\sqrt{5})^2$$ $$4x^2 + 9y^2 = 180$$ 3. Substitute $x^2 = 25-y^2$ $$4(25-y^2) + 9y^2 = 180$$ $$100-4y^2+9y^2=180$$ $$5y^2=80$$ $$y^2=16$$ $$y=4$$ $$x=3$$ Negative root rejected because side lengths are positive.