If one roots of $k{(x-1)}^2=8x-14,k>0,$  is twice the other then the value of k is

Given equation is $k{(x-1)}^2=8x-14,k>0$

It can be written as

$k\left(x^2-2x+1\right)=8x-14$

$\Rightarrow k{x}^2-(2k+8)x+k+14=0$

Let the roots be $\alpha$  and $2\alpha ,$  

then sum of the roots : $\alpha +2\alpha =\frac{2k+8}{k}$

And  product of the roots :   $\alpha \left(2\alpha \right)=\frac{k+14}{k}$

$\Rightarrow \alpha =\frac{2k+8}{3k}$

 and $2{\alpha }^2=\frac{k+14}{k}$

$\Rightarrow 2{\left(\frac{2k+8}{3k}\right)}^2=\frac{k+14}{k}$      $\left(\because \alpha =\frac{2k+8}{3k}\right)$

$\Rightarrow 2\left(\frac{4k^2+64+32k}{9k^2}\right)=\frac{k+14}{k}$

$\Rightarrow 8k^2+64k+128=9k^2+126k$

$\Rightarrow k^2+62k-128=0$

$\Rightarrow (k+64)(k-2)=0\Rightarrow k=-64,2.$

$\therefore k=2\left(\because k>0\right)$