Let p(x) be ax2−bx+c=0 where a,b,c are natural nos. and it has roots lying in the interval (1,2), then find the minimum value of a,b,c individually.

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Ireland Vaughan · Accepted Answer

If you multiply out your polynomial to get ax2−a(α+β)x+aαβ we can see that a can be 1. It cannot be 0 if we take &quot;roots&quot; to indicate there are two of them. If a, α, and β are all positive, b&amp;lt;0 and there is no solution to the problem as stated. If the polynomial is ax2−bx+c we have b=a(α+β) with α,β&amp;gt;1, o the minimum value of b is 2. Similarly, c=aαβ must be at least 2. These conditions are not independent. In fact x2−2x+2=0 has no real roots at all. We need to increase b to get the parabola to hit the axis at all. I don&#039;t find any polynomial that fits the requirement with roots at 1,2 disallowed by the fact that the interval is open. If you close the interval, x2−3x+2=0 has roots at 1, 2.

Let p(x) be \(\displaystyle{a}{x}^{{2}}-{b}{x}+{c}={0}\) where a,b,c are

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