Question

Find Value of K so that equation$(x^2-2x{)}^2-(x^2-2x)+(k+2)=0$ has four real solution

Answer 1

given equation is

${\left(x^2-2x\right)}^2-\left(x^2-2x\right)+\left(k+2\right)=0$---------------------$\left(1\right)$

$put{x}^2-2x=y$------------------------------$\left(2\right)$

$thene{q}^n\left(1\right)willbe$

$y^2-y+\left(k+2\right)=0$------------------------$\left(3\right)$

$\because e{q}^n\left(1\right)$ have is real solution this means $e{q}^n\left(2\right)$ $\&$ $\left(3\right)$ both have two real solution$.$

$i.e.,$ $D=b^2-4ac>0$