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Step-by-Step Solution
Step 1: Write down the given quadratic equation
The given quadratic equation is
$$
(c - 5)x^2 - 2cx + (c - 4) = 0, \quad c \neq 5.
$$
We want to find all integral values of $c$ such that one root lies in $(0,2)$ and the other root lies in $(2,3)$.
Step 2: Evaluate the quadratic at key points
To ensure that a root lies in a specific interval, we check sign changes of the polynomial at the endpoints of that interval. Define
$$
f(x) = (c - 5)x^2 - 2cx + (c - 4).
$$
1. At $x = 0$:
$$
f(0) = c - 4.
$$
2. At $x = 2$:
\[
f(2)
= (c - 5)\,(2)^2 \;-\; 2c\,(2) \;+\; (c - 4)
= (c - 5)\,4 \;-\; 4c \;+\; (c - 4)
= 4c - 20 - 4c + c - 4
= c - 24.
\]
3. At $x = 3$:
\[
f(3)
= (c - 5)\,(3)^2 \;-\; 2c\,(3) \;+\; (c - 4)
= 9(c - 5) - 6c + c - 4
= 9c - 45 - 6c + c - 4
= 4c - 49.
\]
Step 3: Condition for a root in $(0, 2)$
For $f(x)$ to have a root in $(0,2)$, $f(0)$ and $f(2)$ must have opposite signs:
\[
f(0)\,f(2) < 0
\quad\Longrightarrow\quad
(c - 4)\,(c - 24) < 0.
\]
This product is negative if and only if $4 < c < 24$. Since $c$ is integral (and $c \neq 5$), the possibilities from this step are
\[
c \in \{5,6,7, \dots, 23\}
\]
but $c \neq 5$ is given, so effectively
\[
c \in \{6,7,8, \dots, 23\}.
\]
Step 4: Condition for a root in $(2, 3)$
Similarly, for $f(x)$ to have a root in $(2,3)$, $f(2)$ and $f(3)$ must have opposite signs:
\[
f(2)\,f(3) < 0
\quad\Longrightarrow\quad
(c - 24)\,(4c - 49) < 0.
\]
We consider two main sign cases:
1. If $c - 24 > 0 \;(\text{i.e., } c > 24)$, then $4c - 49$ must be $< 0$, which means $4c < 49 \Rightarrow c < 12.25$. This cannot happen simultaneously with $c > 24$. No solutions from this case.
2. If $c - 24 < 0 \;(\text{i.e., } c < 24)$, then $4c - 49$ must be $> 0$, which means $4c > 49 \Rightarrow c > 12.25$. Combining $c > 12.25$ with $c < 24$ gives $12.25 < c < 24$. For integral $c$, this implies
\[
c \in \{13,14,15,16,17,18,19,20,21,22,23\}.
\]
Step 5: Combine both conditions and check discriminant
From Step 3, $c \in \{6,\dots,23\}$. From Step 4, $c \in \{13,14,\dots,23\}$. Their intersection is
\[
c \in \{13,14,15,16,17,18,19,20,21,22,23\}.
\]
That set has 11 elements. We must also ensure the quadratic has real roots, i.e., the discriminant $\Delta \ge 0$.
Discriminant
\[
\Delta = b^2 - 4ac
= (-2c)^2 - 4\,\bigl(c-5\bigr)\,\bigl(c-4\bigr)
= 4c^2 - 4\,(\,c^2 - 9c + 20\,)
= 4c^2 - 4c^2 + 36c - 80
= 36c - 80.
\]
Requiring $\Delta \ge 0$:
\[
36c - 80 \ge 0 \quad\Longrightarrow\quad c \ge \frac{80}{36} = \frac{20}{9}.
\]
All integers $c \ge 3$ satisfy this, so all elements in $\{13,14,\dots,23\}$ certainly do. Hence no further exclusions.
Step 6: Final Answer
Thus, all integral values $c$ in the set $\{13,14,15,16,17,18,19,20,21,22,23\}$ satisfy the given conditions. The number of such values is
$$
11.
$$
Reference Image
