If `16x^4-32x^3+ax^2+bx+1=0`, `a,b in R` has positive real roots only, then `a-b` is equal to :

Options

A. `-32`
B. `32`
C. 49
D. `-49`

Answer

To solve the equation \( 16x^4 - 32x^3 + ax^2 + bx + 1 = 0 \) and find the value of \( a - b \) given that it has only positive real roots, we can follow these steps:

### Step 1: Compare the given polynomial with a known polynomial
We start by rewriting the polynomial in a more manageable form. We can factor out a common term:

\[ 16x^4 - 32x^3 + ax^2 + bx + 1 = 0 \]

### Step 2: Identify a suitable polynomial
We notice that the polynomial resembles the expansion of \( (2x - 1)^4 \). Let's expand \( (2x - 1)^4 \):

\[ (2x - 1)^4 = 16x^4 - 32x^3 + 24x^2 - 8x + 1 \]

### Step 3: Compare coefficients
Now, we can compare the coefficients of our polynomial with the expanded polynomial:

- Coefficient of \( x^2 \): \( a = 24 \)
- Coefficient of \( x \): \( b = -8 \)

### Step 4: Calculate \( a - b \)
Now that we have the values of \( a \) and \( b \):

\[ a - b = 24 - (-8) = 24 + 8 = 32 \]

### Conclusion
Thus, the value of \( a - b \) is:

\[ \boxed{32} \]