Question

Find all values of $n\in N$ for which the equation $\frac{x-8}{n-10}=\frac{n}{x}$ has no solutions.

Answer 1

$\frac{x-8}{x-10}=\frac{x}{x}$

$x^2-8x=x^2-10x$

$x^2-8x-(x^2-10x)=0$

For no solution $D<0.$

So,$64+4(x^2-10x)<0$

$x^2-10x+16<0$

$(x-8)(x-2)<0$

$xϵ(2,8)$

So, $x=3,4,5,6,7$