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    High-Weightage Topics for JEE Main Based on the Last 5 Years

    Location of Roots - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • 23 Questions around this concept.

    Solve by difficulty

    What is the value of $m$ so that both roots of the equation $x^2+m x+1$ are less than unity?

    Let $\mathrm{f}(\mathrm{x})=a x^2+b x+c$, consider the following graph

    Real values of m for which at least one root of $x^2-(m-3) x+m=0$ lies in $(0,1)$ (Given that 0 and 1 are not its roots)

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    Concepts Covered - 3

    Location of roots (1)

    Let f(x) = ax2 + bx + c , where a,b,c are real numbers and ‘a’ is non-zero number. Let ? and ? be the roots of the equation, and let k be a real number. Then:

    1. If both roots of f(x) are less than k then

     

       


     

    i) D ≥ 0 (as the real roots may be distinct or equal)

    ii) af(k) > 0 (In both cases af(k) is positive, as in the second case if a < 0 then f(k) < 0, so multiplying two -ve values will give us a positive value)

    iii)$\mathrm{k}>\frac{-\mathrm{b}}{2 \mathrm{a}}$ since, $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will lie between ? and ?, and ?, ? are less than k so $\frac{-b}{2 a}$ will be less than k.

     

     

    2. If both roots of f(x) are greater than k

     

         



     

    i) D ≥ 0 (as the real roots may be distinct or equal)

    ii) af(k) > 0 (In both cases af(k) is positive, as in the second case if a < 0 then f(k) < 0, so multiplying two -ve values will give us a positive value)

    iii)$\mathrm{k}<\frac{-\mathrm{b}}{2 \mathrm{a}}$ since $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will lie between ? and $\underline{\underline{\underline{?}}}$, and $\underline{\underline{\underline{?}}}$ ? are greater than k so $\frac{-b}{2 a}$ will be greater than k .

    Location of roots (2)

    Condition for number k

    Let f(x) = ax2 + bx + c where a, b, c are real numbers and ‘a’ is a non-zero number. Let ? and ? be the real roots of the function. And let k be any real number. Then:

    1. If k lies between the root ? and ?

     

             

    af(k) < 0.

    (As if a < 0 then f(k) > 0. So multiplying one -ve and one +ve value will give us a negative value)

     

    Condition for numbers k1 and k2

    Let f(x) = ax2 + bx + c, where a,b,c are real numbers and ‘a’ isa  non-zero number. Let ? and ? be the real roots of the function. And let $k_1, k_2$  be any two real numbers. Then:

    2.        If exactly one root of f(x) lies in between the number  $k_1, k_2$

     

       

    $f\left(k_1\right) f\left(k_2\right)<0$  as for one value of k, we will have +ve value of f(x) and for other value of k, we will have a -ve value of f(x) (here  ? < ?)

     

    Location of roots (3)

    Condition on number $k_1, k_2$

    Let f(x) = ax2 + bx + c, where a, b, c are real numbers and ‘a’ is non-zero number. Let ? and ? be the roots of the function. And let $k_1, k_2$  be any two real numbers. Then

     

      1.  If both roots lie between  $k_1, k_2$

           

     

    i) D ≥ 0 (as the real roots may be distinct or equal)

    ii) $\begin{aligned} & \operatorname{af}\left(\mathrm{k}_1\right)>0 \text { and } \mathrm{af}\left(\mathrm{k}_2\right)>0 \\ & \mathrm{k}_1<\frac{-\mathrm{b}}{2 \mathrm{a}}<\mathrm{k}_2, \text { where } \alpha \leq \beta \text { and } \mathrm{k}_1<\mathrm{k}_2\end{aligned}$

     

    2.  If  $k_1, k_2$  lies  between the roots

       

    $\mathrm{af}\left(\mathrm{k}_1\right)<0$ and $\mathrm{af}\left(\mathrm{k}_2\right)<0$  

    Study it with Videos

    Location of roots (1)
    Location of roots (2)
    Location of roots (3)

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