Question 123

If $$(x - \alpha)$$ and $$(x - \beta )$$ are the two factors of the polynomial $$f(x) = ax^{2} + bx + c$$,then the quadratic polynomial whose factors are $$(x - \frac{1}{\alpha})$$ and $$(x - \frac{1}{\beta})$$ is

Solution

Solution:

If factors are (x- α) and (x - β) then roots of the expression will be α , β.

Product of roots = α x β = (c/a) ..........Eqn (i)

Sum of roots = α + β = -(b/a)  ..........Eqn (ii)

In the new case,

Let polynomial be Mx² + NX + O

when factors are $$(x - \frac{1}{\alpha})$$ and $$(x - \frac{1}{\beta})$$ 

Roots will be $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$ 

So product of roots = O/M = (1/α) x (1/β) = (1/αβ) = a/c …….from equation (i)

Sum of roots = -N/M = (1/α) + (1/β) = {(α + β) / αβ} = (-b/a) / (c/a) = - b/c…….from equation (ii)

So product of roots = a/c = O/M

Sum of roots = - b/c = - N/M

On comparison, the polynomial Mx2 + NX + O turns out to be

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