Question

Find the value of $a$ so that the equation $f\left(x\right)=x^2+(a-3)x+a=0$ has exactly one root $\alpha$, between the interval $(1,2)$ and $f(x+\alpha )=0$ has exactly one root between the interval $(0,1)$.

Answer 1

REF.Image.

From graph

$f\left(1\right).l\left(2\right)<0$

$(1+a-3+a)(4+2a-6+a)<0$

$(2a-2)(3a-2)<0$

$\Rightarrow 2(a-1)(3a-2)<0$

$\Rightarrow (a-1)(3a-2)<0$

$\therefore aϵ(\frac{2}{3},1)$ __ (i)

$f(x+\alpha )=(x+\alpha {)}^2+(a-3\left)\right(x+\alpha )+a$

$\therefore f\left(0\right)f\left(1\right)<0$

$a.(2a-2)<0$

$2a(a-1)<0$

$\therefore aϵ(0,1)$ __ (ii)

From equation (i) & (ii)

$aϵ(\frac{2}{3},1)$