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Question: If the both roots of the quadratic equation \({{x}^{2}}-mx+4=0\) are real and distinct and they lie ...

If the both roots of the quadratic equation x2mx+4=0{{x}^{2}}-mx+4=0 are real and distinct and they lie in the interval [1,5], then m lies in the interval:
(A) (4,5) (B) (3,4) (C) (5,6) (D) (5,4) \begin{aligned} & \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{aligned}

Explanation

Solution

We solve this question by going through the nature of roots of a quadratic equation using discriminant Δ=b24ac\Delta ={{b}^{2}}-4ac and find the range when the discriminant is greater than zero.
Then we take the sum of roots using formula α+β=ba\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 ax2+bx+c=0a{{x}^{2}}+bx+c=0 having roots α\alpha and β\beta ,
Sum of the roots is α+β=ba\alpha +\beta =\dfrac{-b}{a} and product of the roots is αβ=ca\alpha \beta =\dfrac{c}{a}.
The nature of the roots can be said by the discriminant of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0, that is Δ=b24ac\Delta ={{b}^{2}}-4ac.
If the roots are real and distinct, then the discriminant is greater than zero, which is b24ac>0{{b}^{2}}-4ac>0.
If the roots are equal, then the discriminant is equal to zero, which is b24ac=0{{b}^{2}}-4ac=0.
If the roots are imaginary, then the discriminant is less than zero, which is b24ac<0{{b}^{2}}-4ac<0.
We were given that roots of the equation x2mx+4=0{{x}^{2}}-mx+4=0 are distinct.
So, from above as the roots are real and distinct, its discriminant is greater than zero.
Δ=(m)24(1)(4)>0 m216>0 (m4)(m+4)>0 m(,4)(4,)............(1) \begin{aligned} & \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{aligned}
So, we get that the range of m as above.
Let the roots of given equation be α\alpha and β\beta .
So, by using the formula for sum of the roots of a quadratic equation, we have
α+β=(m) α+β=m \begin{aligned} & \Rightarrow \alpha +\beta =-\left( -m \right) \\\ & \Rightarrow \alpha +\beta =m \\\ \end{aligned}.
We were also given that the roots of the given quadratic equation lie between 1 and 5, that is
1<α<5 1<β<5 \begin{aligned} & \Rightarrow 1<\alpha <5 \\\ & \Rightarrow 1<\beta < 5 \\\ \end{aligned}
Adding the above two inequalities we get,
1+1<α+β<5+5 2<α+β<10 2<m<10.............(2) \begin{aligned} & \Rightarrow 1+1< \alpha +\beta < 5+5 \\\ & \Rightarrow 2< \alpha +\beta < 10 \\\ & \Rightarrow 2< m <10.............\left( 2 \right) \\\ \end{aligned}
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,
(1)2m(1)+4>0 1m+4>0 5m>0 m<5..............(3) \begin{aligned} & \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{aligned}
When x=5,
(5)2m(5)+4>0 255m+4>0 295m>0 5m<29 m<295..............(4) \begin{aligned} & \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{aligned}
From equations (1), (2), (3) and (4) we can find the region common to all of them. So the region common to m(,4)(4,)m\in \left( -\infty ,-4 \right)\cup \left( 4,\infty \right), 2<m<102< m< 10, m<5m< 5, m<295m< \dfrac{29}{5} is m(4,5)m\in \left( 4,5 \right).
So, m lies in the interval (4,5)\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(,4)(4,)m\in \left( -\infty ,-4 \right)\cup \left( 4,\infty \right) and 2<m<102< m< 10, and the region common to them is (4,10)\left( 4,10 \right). But it is wrong. So, one should consider all the possible ways of finding the interval.