Question

If ${ax}^2-bx+c=0$ have two distinct roots lying in the interval $(0,1),a,b,c\in N$, then find the value of ${\log }_{s}abc$?

Answer 1

We need to show that $abc\ge 25$.

Since both roots are real and distinct, we have that $b^2-4ac>0$ and so $b^2>4ac$.

Since both roots are in (0,1), their average $\frac{b}{2a}<1$ and therefore $b<2a$.

Since the larger root $\frac{b+\sqrt{b^2-4ac}}{2a}<1$, we have that $b+\sqrt{b^2-4ac}<2a$ and therefore $\sqrt{b^2-4ac}<2a-b$.

Squaring both sides gives $b^2-4ac<4a^2-4ab+b^2$, so $4ab<4a^2+4ac$ and therefore b<a+c.

Since 2a>b, we have that $4a^2>b^2>4ac$ and therefore a>c.

Since $b^2\ge 4ac+1$ and $a\ge c+1$, we conclude that $b^2\ge 4c(c+1)+1=(2c+1{)}^2$ and thus $b\ge 2c+1$.

Therefore $abc\ge (c+1)(2c+1)c>25$ if $c\ge 2$.

When c=1, b<a+1 implies that $b\le a$, so $a^2\ge b^2>4a$ and therefore $a\ge 5$.

Then $b^2>4a\ge 20$, so $b\ge 5$ and $abc=ab\ge 25$.