Write $x^4-2ax^3+a(a+1)x^2-2ax+a^2=0$ as $(x^2+1)(x-a)^2 = (1-a)x^2$. Therefore in order to have a real root we need $a\leq 1$, and indeed, for $a=1$ the equation becomes $(x^2+1)(x-1)^2 = 0$, with $0$ as double root. For any $a<1$, comparing the graphs of $(1-a)x^2$ and $(x^2+1)(x-a)^2$ it is seen they must intersect.

For part b), we can also use the fact $f(a)=a^3-a^2\le0$ when $a<1$.

Thnx guys

mavropnevma wrote: Write $x^4-2ax^3+a(a+1)x^2-2ax+a^2=0$ as $(x^2+1)(x-a)^2 = (1-a)x^2$. Can I ask the motivation behind writing this? @below I couldn't understand your point.Please explain!

Perfect squares are non-negative.

ayan_mathematics_king wrote: mavropnevma wrote: Write $x^4-2ax^3+a(a+1)x^2-2ax+a^2=0$ as $(x^2+1)(x-a)^2 = (1-a)x^2$. Can I ask the motivation behind writing this? @below I couldn't understand your point.Please explain! YES what was the motivation ?? i always meet some weird factorisations but from where people come up with those weird ones i dk ?

I'd say the main motivation are those factors $2$. One of the main ways to discover apparently impossible factorisations is to look for perfect squares and add/subtract lacking terms, adjusting what is left. In this case, the factors "2" remind us about the square of sum. Observing the expression $-2ax^3$, plus the fact there's just $x^4$ with higher degree, we try $(x^2 - ax)^2 = (x^2)^2 - 2(x^2)(ax) + (ax)^2$. Thus, $x^4 - 2ax^3 + a(a + 1)x^2 - 2ax + a^2 = (x^2 - ax)^2 + ax^2 - 2ax + a^2 = x^2(x - a)^2 + ax^2 - 2ax + a^2$. Now, observing the expression $-2ax$, plus the fact there's just $a^2$ with lower degree, we try $(x - a)^2 = x^2 - 2ax + a^2$. Thus, $x^2(x - a)^2 + ax^2 - 2ax + a^2 = x^2(x - a)^2 + (a - 1)x^2 + (x - a)^2 = (x^2 + 1)(x - a)^2 + (a - 1)x^2$. Finally, $x^4 - 2ax^3 + a(a + 1)x^2 - 2ax + a^2 = 0 <=> (x^2 + 1)(x - a)^2 + (a - 1)x^2 = 0 <=> (x^2 + 1)(x - a)^2 = (1 - a)x^2$.