Let $a,b\in R$ such that the equation $a{x}^2-2bx+15=0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation $x^2-2bx+21=0$ then ${\alpha }^2+{\beta }^2$ is equal to:

$$\begin{gathered} a{x}^2-2bx+15=0 has repeated roots \alpha ,\alpha \\ \Rightarrow 2\alpha =\frac{2b}{a},{\alpha }^2=\frac{15}{a} \\ \Rightarrow \frac{\alpha }{2}=\frac{15}{2b} \\ \alpha =\frac{15}{b} \\ \sin ce \alpha is the root of the equation x^2-2bx+21=0 then \\ {\left(\frac{15}{b}\right)}^2-2b\left(\frac{15}{b}\right)+21=0 \\ {b}^2=25 \\ \alpha +\beta =2b,\alpha \beta =21 \\ {\alpha }^2+{\beta }^2=4b^2-42 \\ =58 \end{gathered}$$