If one root of equation `(l-m) x^2 + lx + 1` = 0 be double of the other and if `l` be real, show that `m<=9/8`

Options

A. `9/8`
B. `7/8`
C. `8/9`
D. `5/9`

Answer

To solve the problem, we need to show that if one root of the equation \((l - m)x^2 + lx + 1 = 0\) is double the other, then \(m \leq \frac{9}{8}\).

### Step-by-Step Solution:

1. **Assume the Roots**:
Let the roots of the equation be \(\alpha\) and \(2\alpha\) (since one root is double the other).

2. **Sum of Roots**:
From Vieta's formulas, the sum of the roots \(\alpha + 2\alpha = 3\alpha\) is given by:
\[ 3\alpha = -\frac{b}{a} = -\frac{l}{l - m} \]
Therefore, we can express \(\alpha\) as:
\[ \alpha = -\frac{l}{3(l - m)} \]

3. **Product of Roots**:
The product of the roots is given by:
\[ \alpha \cdot 2\alpha = 2\alpha^2 = \frac{c}{a} = \frac{1}{l - m} \]
Substituting \(\alpha\) into this equation gives:
\[ 2\left(-\frac{l}{3(l - m)}\right)^2 = \frac{1}{l - m} \]
Simplifying this, we have:
\[ 2 \cdot \frac{l^2}{9(l - m)^2} = \frac{1}{l - m} \]

4. **Cross-Multiplying**:
Cross-multiplying yields:
\[ 2l^2 = 9(l - m) \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ 2l^2 - 9l + 9m = 0 \]

6. **Discriminant Condition**:
For \(l\) to be real, the discriminant of the quadratic equation must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \(a = 2\), \(b = -9\), and \(c = 9m\):
\[ (-9)^2 - 4 \cdot 2 \cdot 9m \geq 0 \]
Simplifying this gives:
\[ 81 - 72m \geq 0 \]

7. **Solving the Inequality**:
Rearranging the inequality:
\[ 81 \geq 72m \]
Dividing both sides by 72:
\[ m \leq \frac{81}{72} = \frac{9}{8} \]

### Conclusion:
Thus, we have shown that if one root of the equation \((l - m)x^2 + lx + 1 = 0\) is double the other and \(l\) is real, then \(m \leq \frac{9}{8}\).