It's quadratic again

Algebra Level 3

A quadratic polynomial f ( x ) = a x 2 + b x + c f(x)=ax^2+bx+c has α \alpha and β \beta as its roots, where 2 < β < 3 2<\beta<3 , a c < 0 ac<0 and b − a c > 1 \dfrac {b-a}c > 1 . Then what is the range of α \alpha ?

-3<α<0 0<α<1 -1<α<0 -3<α<-2

Durgaprasad Sahu by durgaprasad sahu , 17, India

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1 solution

It's given that 1 of the root will lye between 2&3.

Since ac<0 , there will be definitely 2 roots. (Quadratic equation)

Assume a>0, (parabola is concave up)

  1. At X=0 , fn is -ve Hence second root must be before 0.

  2. From X=-1 and the inequality , fn is +ve . Therefore, root must be greater than -1.

Same result for a<0 ... i.e. -1< alpha<0

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