Question

If both the roots of the quadratic equation x² - 5x + 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)

Answer 1

answer : option (A) (4,5)

given, If both the roots of the quadratic equation x² - mx + 4 = 0 are real and distinct and they lie in the interval [1, 5].

discriminant, D = b² - 4ac > 0 for real and distinct.

= (-m)² - 4 × 4 × 1 > 0

= m² - 16 > 0

= (m - 4)(m + 4) > 0

m > 4 or m < -4 ...........(1)

again, f(x) = x² - mx + 4

when roots lie in the interval [1, 5].

f(1) = 1 - m + 4 = 5 - m > 0 ⇒m < 5 .....(2)

f(5) = 25 - 5m + 4 > 0 ⇒m < 29/5 .....(3)

and 1 ≤ -b/2a ≤ 5 ⇒ 1 ≤ -(-m)/2 ≤ 5

⇒2 ≤ m ≤ 10 .......(4)

putting all equations in number line,

we get, 4 < m < 5 , hence option (A) is correct choice.

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Answer 2

Option A: The number m lies in the interval $(4,5)$

Explanation:

Given that the roots of the quadratic equation $x^{2}-m x+4=0$ are real and distinct and they lie in the interval $[1,5]$.

We need to determine the interval in which m lies.

Using the discriminant formula, we have,

$b^2-4ac>0$

Substituting the values, we get,

$m^2-4(1)(4)>0$

       $m^2-16>0$

               $m^2>16$

               $m>\pm4$

Thus, m lies in the $m \in(-\infty,-4) \cup(4, \infty)$

Let us substitute $x=1$ in the equation $x^{2}-m x+4=0$

Thus, we have,

$f(1) \geq 0 \Rightarrow 5-m \geq 0$

This implies that $m \in(-\infty, 5]$

Also, let us substitute $x=5$, we get,

$f(5) \geq 0 \Rightarrow 29-5 m \geq 0$

This implies that $m \in\left(-\infty, \frac{29}{5}\right]$

Since, the roots lie in the interval $[1,5]$, we have,

$1<\frac{-b}{2 a}<5$

$1<\frac{m}{2}<5$

Thus, we get,

$m \in(2,10)$

Thus, the interval in which m lies is $(4,5)$

Therefore, Option A is the correct answer.

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