Math Archives (2)

$(1). \text{Reciprocal roots}\implies c/a=1\implies c=a$ $(2). \text{Distinct roots} \implies b^2-4ac=b^2-4a^2>0\\\implies \dfrac{b^2}{a^2}-4>0~(\because a\neq 0)$ $(3).\text{positive roots } \implies \dfrac{b}{a}<0$

Substituting $b/a=t$ , From $(2).$ and $(3).$ , it can be easily established that $\boxed{t<-2}$

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$\begin{aligned}af'(1)=&a(2a+b)\\=&a^2(2+\dfrac{b}{a})\\=&a^2(2+t)<0\end{aligned}$

$(\because t<-2 \text{ and } a^2>0)$

Let $a$ and $1/a$ be the positive roots, then $f(x) = (x-a)(x-\frac 1 a) = x^2 - (a+\frac 1 a)x + 1. \\ f'(x) = 2x - (a + \frac 1 a). \\ f'(1) = 2 - a - \frac 1a. \\ a\:f'(1) = 2a - a^2 - 1 = -(a-1)^2.$ Since $a \not= 0$ , the last expression is certainly $\boxed{< 0}$ .

Let roots of the given polynomial be x1,x2

As the roots of the given equation are distinct and reciprocal of each other

Hence x1*x2 = 1 = c/a

Thus c= a ; Now the polynomial becomes f(x) = ax^2 + bx + a

f(1) = 2a + b ---------------------------(1)

f ' (x) = 2ax + b

f'(1) = 2a + b = f(1) { From equation (1)}

Note: x1 , x2 not equal to 1 { As they are distinct }

From above it can be easily proved that at least one of the roots of f(x) has to be greater than one and the other root less than 1

Case 1: a < 0

In this case from graph of a quadratic polynomial f'(1) = f(1) > 0 as { x2 < 1 <x1}

Thus a*f'(1) < 0

Case 2: a > 0

Again using graph of a quadratic polynomial with leading coefficient positive

f'(1) = f(1) < 0 { x2 <1 < x1}

Thus a*f'(1) < 0