If $$x(2x+3)=90$$ and $$7y^{\frac{-1}{2}}+2y^{\frac{-1}{2}}=y^{\frac{1}{2}}$$ (x and y are positive numbers), then what is the value of $$x^2 + y^2$$ ?

Given : $$x(2x+3)=90$$

=> $$2x^2+3x-90=0$$

=> $$2x^2-12x+15x-90=0$$

=> $$2x(x-6)+15(x-6)=0$$

=> $$(x-6)(2x+15)=0$$

=> $$x=6,\frac{-15}{2}$$

$$\because x$$ is positive, => $$x=6$$ ---------(i)

Also, $$7y^{\frac{-1}{2}}+2y^{\frac{-1}{2}}=y^{\frac{1}{2}}$$

=> $$\frac{7}{\sqrt{y}}+\frac{2}{\sqrt{y}}=\sqrt{y}$$

=> $$\frac{9}{\sqrt{y}}=\sqrt{y}$$

=> $$y=9$$ --------(ii)

$$\therefore$$ $$x^2+y^2$$

= $$(6)^2+(9)^2=36+81=117$$

=> Ans - (C)