Question

If ${ax}^2-bx+c=0$ has two distinct roots lying in the interval $(0,1)$ and $a,b,c\in N$, then ${\log }_{5}abc\ge k$. The value of $k$ is

Answer 1

Given that the two roots of $a{x}^2-bx+c=0$ lies in (0,1) $a,b,cϵN$

let $f\left(x\right)=a{x}^2-bx+cf\left(0\right)>0andf\left(1\right)>0\therefore c>0anda-b+c>0$

Minimum value of f(x) is at x in between (0,1)

$\therefore \frac{-(-b)}{2a}<1b<2a$

there are two distinct roots

$\therefore b^2-4ac>0b^2>4ac\Rightarrow b^2\ge 4ac+12a>b\Rightarrow 4a^2>b^2\Rightarrow 4a^2>4ac(\because b^2>4ac)\therefore a>c\therefore a\ge c+1b^2\ge 4ac+1\Rightarrow b^2\ge 4c(c+1)+1\Rightarrow b^2\ge (2c+1{)}^2\therefore b\ge 2c+1(\because a,b,cϵN)\therefore abc\ge c(c+1\left)\right(2c+1)ifc=1,thatmeansa-b+c>0\Rightarrow b<a+1\therefore a\ge b{a}^2\ge b^2>4a(\because b^2>4ac,c=1)\therefore a^2>4a\Rightarrow a>4\Rightarrow a\ge 5b^2>4ac\Rightarrow b^2>4ac\Rightarrow b^2>20(\because a\ge 5)\therefore b\ge 5(\because bϵN)\therefore abc=ab\ge 25(\because c=1)ifc\ge 2,thenc(c+1\left)\right(2c+1)\ge 2(2+1\left)\right(5)\ge 30\therefore ifc\ge 1i.e.ifcϵN,abc\ge 25\therefore lo{g}_{5}abc\ge 2$