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Location of Roots - Practice Questions & MCQ

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

Quick Facts

  • 26 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

Let $a_1<a_2<a_3<a_4$ then the number of real roots of equation $\left(x-a_1\right)\left(x-a_3\right)+\left(x-a_2\right)\left(x-a_4\right)=0$ equals

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)

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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