The roots of the quadratic equation lie between some range and it depends upon the coefficients of the equation. or may be predicted using the relation between coefficients.
Try the problem
Suppose \(a+b+c\) and \(a-b+c\) are positive and \(c<0\). Then the equation \(ax^2+bx+c=0\)
(A) has exactly one root lying between -1 and +1
(B) has both the roots lying between -1 and +1
(C) has no root lying between -1 and +1
(D) nothing definite can be said about the roots without knowing the value of a,b and c.
ISI entrance B. Stat. (Hons.) 2003 problem 4
6 out of 10
challenges and thrills of pre college mathematics
Use some hints
The question is based upon mean value theorem and the lower and upper limits are 1 and -1
So let \(f(x)=ax^2+bx+c\) now find f(1) and f(-1), and try to use mean value theorem.
So \(f(1)=a+b+c\) and \(f(-1)=a-b+c\) and we can see that both of them are positive. so now using mean value theorem we can find c in \(f'(c)\).
An now we can apply Sridharacharya’s formula to find the valise(s) of x.
We know c in f'(c) represent the roots and the mean value theorem says c must lie between limits (-1 and +1 in this case)
So option B is correct one.
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