I have to find the values of $k$ for which the equation:$$x^4-(k-3)x^2+k=0$$ has : (i) four real roots and (ii) exactly 2 real roots.
My solution: put $x^2=t$, then the equation becomes $$t^2-(k-3)t+k=0 ......(1)$$ For (i) we need (1) to have 2 positive roots-real or distinct. For (ii) we need (1) to have one positive and one negative root.
But what my book does: For (i) it only considers the case when (1) has positive and distinct roots, and hence doesn't consider k=9 in the solution set. But I think that $(x^2-3)^2=0$ has 4 real roots.
For (ii), in addition to my consideration, it also takes into account the case where (1) has real equal positive roots. But that would make the original equation to have 4 real roots.
Also the book mentioned a few pages ago that:
Difference between root and solution: A root of a polynomial equation may be real or imaginery while a solution has to be real. A quadratic equation having two distinct real roots is said to have two solutions. If it has two equal real roots then it has one solution. If it has two imaginery roots then we say that the equation has no solution.
Is the book contradicting its own(extrapolated) statement or am I wrong?
My solution for (i):
for (1) to have two positive roots(equal or distinct) we need:
Sum of roots$\gt0$ that is $k-3>0\Rightarrow k>3$,
Product of roots$\gt0$ that is $k>0$ and
$$D\geq0$$
$$(k-1)(k-9)\geq0$$ $$k\in(-\infty,1]\cup[9,\infty)$$
Taking intersection of all these intervals we get$$[9,\infty)$$.
This has 4 real roots but those roots are pairwise equal. This means that the roots are $\sqrt{3},\sqrt{3},-\sqrt{3},-\sqrt{3}$. Your book probably means 4 distinct real roots, which this equation doesn't have.
For (ii), as you said, if (1) has a real equal positive root $u$, then $x^2 = u \implies x = \pm \sqrt{u}$. This also has pairwise equal roots $ \sqrt{u}, \sqrt{u}, - \sqrt{u}, - \sqrt{u}$. This means that it would have 4 roots in this case.
Your book probably missed out the word distinct in both parts of the question. Adding that word would make the textbook answer the correct one.