Question

If both roots of the equation $x^2 - 5ax + 6a = 0$ exceed 1, then the range of 'a' is

• A. $[-1, 0) \cup [\frac{24}{25}, \infty)$
• B. $[\frac{24}{25}, \infty)$
• C. $[-1, 0)$
• D. $\mathbb{R}$

Hint:

Problems on the location of roots always involve a combination of three conditions: the discriminant ($D$), the value of the function at the given point ($f(k)$), and the position of the vertex ($-B/2A$). It's helpful to visualize the parabola to understand why each condition is necessary. Drawing a number line to find the intersection of the intervals obtained from each condition can prevent errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a quadratic equation $Ax^2+Bx+C=0$ with real roots, the location of the roots relative to a certain number $k$ can be determined by a set of conditions. Here, we need both roots to be greater than 1.
Step 2: Key Formula or Approach
Let $f(x) = x^2 - 5ax + 6a$. For both roots of $f(x)=0$ to be greater than a number $k$ (here $k=1$), the following three conditions must be satisfied:
1. The discriminant must be non-negative: $D = B^2 - 4AC \ge 0$.
2. The value of the function at $k$ must have the same sign as the leading coefficient $A$: $A \cdot f(k)>0$.
3. The x-coordinate of the vertex must be greater than $k$: $-\frac{B}{2A}>k$.
Step 3: Detailed Explanation
Here, $A=1$, $B=-5a$, $C=6a$, and $k=1$.
Condition 1: $D \ge 0$
\[ (-5a)^2 - 4(1)(6a) \ge 0 \] \[ 25a^2 - 24a \ge 0 \] \[ a(25a - 24) \ge 0 \]
This inequality holds when $a \le 0$ or $a \ge \frac{24}{25}$.
So, $a \in (-\infty, 0] \cup [\frac{24}{25}, \infty)$.
Condition 2: $A \cdot f(1)>0$
Since $A=1>0$, we need $f(1)>0$.
\[ f(1) = (1)^2 - 5a(1) + 6a>0 \] \[ 1 - 5a + 6a>0 \] \[ 1 + a>0 \implies a>-1 \] So, $a \in (-1, \infty)$.
Condition 3: $-\frac{B}{2A}>1$
\[ -\frac{-5a}{2(1)}>1 \] \[ \frac{5a}{2}>1 \] \[ 5a>2 \implies a>\frac{2}{5} \] So, $a \in (\frac{2}{5}, \infty)$.
Now, we need to find the intersection of the solution sets from all three conditions.
From (1): $a \in (-\infty, 0] \cup [\frac{24}{25}, \infty)$
From (2): $a \in (-1, \infty)$
From (3): $a \in (\frac{2}{5}, \infty)$
Let's find the intersection.
Intersection of (2) and (3): $(-1, \infty) \cap (\frac{2}{5}, \infty) = (\frac{2}{5}, \infty)$.
Now, intersect this with (1):
\[ \left( (-\infty, 0] \cup [\frac{24}{25}, \infty) \right) \cap \left(\frac{2}{5}, \infty\right) \] We compare the values: $\frac{2}{5} = 0.4$ and $\frac{24}{25} = 0.96$.
The interval $(\frac{2}{5}, \infty)$ does not overlap with $(-\infty, 0]$.
We need to find the intersection of $[\frac{24}{25}, \infty)$ and $(\frac{2}{5}, \infty)$.
Since $\frac{24}{25}>\frac{2}{5}$, the intersection is $[\frac{24}{25}, \infty)$.
Step 4: Final Answer
The range of values for 'a' that satisfies all three conditions is $[\frac{24}{25}, \infty)$.