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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.

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.

Source

Competency

Difficulty

Suggested Book

ISI entrance B. Stat. (Hons.) 2003 problem 4

Quadratic equation

6 out of 10

challenges and thrills of pre college mathematics

First hint

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.

Second Hint

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.

Final Step

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.

- https://www.cheenta.com/problem-based-on-divisibility-cmi-2015-problem-3/
- https://www.youtube.com/watch?v=P4ZYA4XCQoM

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