If one root of `2x^(2)-5x+k=0` be double the other, find the value of k.

Answer

To solve the problem where one root of the quadratic equation \(2x^2 - 5x + k = 0\) is double the other, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the equation be \(\alpha\) and \(2\alpha\).

### Step 2: Use the Sum of Roots Formula
According to Vieta's formulas, the sum of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ \text{Sum of roots} = -\frac{b}{a} \]
For our equation, \(a = 2\) and \(b = -5\). Thus, we have:
\[ \alpha + 2\alpha = -\frac{-5}{2} \]
This simplifies to:
\[ 3\alpha = \frac{5}{2} \]
From this, we can solve for \(\alpha\):
\[ \alpha = \frac{5}{6} \]

### Step 3: Use the Product of Roots Formula
The product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
For our equation, this means:
\[ \alpha \cdot (2\alpha) = \frac{k}{2} \]
Substituting \(\alpha\) into the equation:
\[ \alpha \cdot 2\alpha = 2\alpha^2 = \frac{k}{2} \]
Now substituting \(\alpha = \frac{5}{6}\):
\[ 2\left(\frac{5}{6}\right)^2 = \frac{k}{2} \]
Calculating \(\left(\frac{5}{6}\right)^2\):
\[ \left(\frac{5}{6}\right)^2 = \frac{25}{36} \]
Thus:
\[ 2 \cdot \frac{25}{36} = \frac{k}{2} \]
This simplifies to:
\[ \frac{50}{36} = \frac{k}{2} \]
Multiplying both sides by 2:
\[ \frac{100}{36} = k \]
Simplifying \(\frac{100}{36}\):
\[ k = \frac{25}{9} \]

### Conclusion
The value of \(k\) is \(\frac{25}{9}\).

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