Question

In the following equation, find the value of $\frac{k}{ab}$ for which the given value is a solution of the given equation : $x^2-x(a+b)+k=0,Givenvaluex=a$

Answer 1

Given,

$x^2-x(a+b)+k=0$

$x=a$

$\Rightarrow a^2-a(a+b)+k=0$

$a^2-a^2-ab+k=0$

$\therefore k=ab$

Given,

$\frac{k}{ab}=\frac{ab}{ab}=1$