Question

Show that there is no real number p for which the equation $x^2-3x+p=0$ has two distinct roots in [0,1].

Answer 1

Let, if possible, there are two distinct roots $a$ and $b$ of the given equation in $(0,1)$, such that $0<a<b<1$
Now , let $f\left(x\right)=x^2-3x-p$
Obviously, $f\left(x\right)$ is continous and differentiable for all values of $x$ (being a polynomial)
Also, we have $f\left(a\right)=f\left(b\right)=0$
$f\left(x\right)$ satisfies all the conditions of Rolle's theorem in $(a,b)$ hence $\exists$ a point $cϵ[a,b]$ such that $f^{{}^{\prime }}\left(c\right)-0,$

Now $f^{{}^{\prime }}\left(x\right)=0\Rightarrow 3x^2-3=0\Rightarrow =x=\pm 1$ which is a contradiction [\because $a<c<b$ as $0<a<b<1$]
$\Rightarrow$ our assumption is wrong .
Hence, there can not be two distinct roots of $f\left(x\right)=0,x\in [0,1]$ for any value of $p$