Question

If the both roots of the quadratic equation ${{x}^{2}}-mx+4=0$ are real and distinct and they lie in the interval [1,5], then m lies in the interval:
$\begin{align}
  & \left( A \right)\text{ }\left( 4,5 \right) \\
 & \left( B \right)\text{ }\left( 3,4 \right) \\
 & \left( C \right)\text{ }\left( 5,6 \right) \\
 & \left( D \right)\text{ }\left( -5,-4 \right) \\
\end{align}$

Answer 1

Hint: We solve this question by going through the nature of roots of a quadratic equation using discriminant $\Delta ={{b}^{2}}-4ac$ and find the range when the discriminant is greater than zero.
Then we take the sum of roots using formula $\alpha +\beta =\dfrac{-b}{a}$ and find another range of m. Then we plot the rough graph from the given information and find the range of m when x=1 and x=5 by substituting those values in the equation and by inequality that their values are greater than zero.

Complete step-by-step solution:
First, let us start by going through the concept of the nature of roots before starting to solve the question.
For any quadratic equation $a{{x}^{2}}+bx+c=0$ having roots $\alpha $ and $\beta $,
Sum of the roots is $\alpha +\beta =\dfrac{-b}{a}$ and product of the roots is $\alpha \beta =\dfrac{c}{a}$.
The nature of the roots can be said by the discriminant of the quadratic equation $a{{x}^{2}}+bx+c=0$, that is $\Delta ={{b}^{2}}-4ac$.
If the roots are real and distinct, then the discriminant is greater than zero, which is ${{b}^{2}}-4ac>0$.
If the roots are equal, then the discriminant is equal to zero, which is ${{b}^{2}}-4ac=0$.
If the roots are imaginary, then the discriminant is less than zero, which is ${{b}^{2}}-4ac<0$.
We were given that roots of the equation ${{x}^{2}}-mx+4=0$ are distinct.
So, from above as the roots are real and distinct, its discriminant is greater than zero.
$\begin{align}
  & \Rightarrow \Delta ={{\left( -m \right)}^{2}}-4\left( 1 \right)\left( 4 \right)>0 \\
 & \Rightarrow {{m}^{2}}-16>0 \\
 & \Rightarrow \left( m-4 \right)\left( m+4 \right)>0 \\
 & \Rightarrow m\in \left( -\infty ,-4 \right)\cup \left( 4,\infty \right)............\left( 1 \right) \\
\end{align}$
So, we get that the range of m as above.
Let the roots of given equation be $\alpha $ and $\beta $.
So, by using the formula for sum of the roots of a quadratic equation, we have
$\begin{align}
  & \Rightarrow \alpha +\beta =-\left( -m \right) \\
 & \Rightarrow \alpha +\beta =m \\
\end{align}$.
We were also given that the roots of the given quadratic equation lie between 1 and 5, that is
$\begin{align}
  & \Rightarrow 1<\alpha <5 \\
 & \Rightarrow 1<\beta < 5 \\
\end{align}$
Adding the above two inequalities we get,
$\begin{align}
  & \Rightarrow 1+1< \alpha +\beta < 5+5 \\
 & \Rightarrow 2< \alpha +\beta < 10 \\
 & \Rightarrow 2< m <10.............\left( 2 \right) \\
\end{align}$
Now let us plot the graph of the quadratic equation using the given information.

As we see in the graph, when x=1 and when x=5 the value of the equation is positive. So,
When x=1,
$\begin{align}
  & \Rightarrow {{\left( 1 \right)}^{2}}-m\left( 1 \right)+4>0 \\
 & \Rightarrow 1-m+4 > 0 \\
 & \Rightarrow 5-m >0 \\
 & \Rightarrow m < 5..............\left( 3 \right) \\
\end{align}$
When x=5,
$\begin{align}
  & \Rightarrow {{\left( 5 \right)}^{2}}-m\left( 5 \right)+4>0 \\
 & \Rightarrow 25-5m+4 >0 \\
 & \Rightarrow 29-5m> 0 \\
 & \Rightarrow 5m< 29 \\
 & \Rightarrow m< \dfrac{29}{5}..............\left( 4 \right) \\
\end{align}$
From equations (1), (2), (3) and (4) we can find the region common to all of them. So the region common to $m\in \left( -\infty ,-4 \right)\cup \left( 4,\infty \right)$, $2< m< 10$, $m< 5$, $m< \dfrac{29}{5}$ is $m\in \left( 4,5 \right)$.
So, m lies in the interval $\left( 4,5 \right)$.
Hence, the answer is Option A.

ote: The major mistake that one does in this question is while finding the intervals they forget to check the condition that the equation is positive when x=1 and when x=5. In that case they have only two equations of intervals $m\in \left( -\infty ,-4 \right)\cup \left( 4,\infty \right)$ and $2< m< 10$, and the region common to them is $\left( 4,10 \right)$. But it is wrong. So, one should consider all the possible ways of finding the interval.