IF ` x^2 +bx +a=0, ax^2+x+b=0` have a common root and the first equation has equal roots then ` 2a^2+b=`

Options

A. 0
B. -1
C. 1
D. 2

Answer

To solve the problem, we need to analyze the two quadratic equations given and use the information provided about their roots.

### Step 1: Identify the equations
The two equations given are:
1. \( x^2 + bx + a = 0 \)
2. \( ax^2 + x + b = 0 \)

### Step 2: Conditions for the first equation
The first equation has equal roots. For a quadratic equation \( Ax^2 + Bx + C = 0 \) to have equal roots, the discriminant must be zero:
\[ D_1 = b^2 - 4a = 0 \]
From this, we can express \( b^2 = 4a \).

### Step 3: Common root condition
Let \( r \) be the common root of both equations. Substituting \( r \) into the first equation gives:
\[ r^2 + br + a = 0 \quad (1) \]
Substituting \( r \) into the second equation gives:
\[ ar^2 + r + b = 0 \quad (2) \]

### Step 4: Substitute \( r^2 \) from equation (1) into equation (2)
From equation (1), we can express \( r^2 \):
\[ r^2 = -br - a \]
Substituting this into equation (2):
\[ a(-br - a) + r + b = 0 \]
Expanding this gives:
\[ -ab r - a^2 + r + b = 0 \]
Rearranging terms:
\[ (-ab + 1)r + (b - a^2) = 0 \]

### Step 5: Solve for \( r \)
For the above equation to hold for all \( r \), both coefficients must equal zero:
1. \( -ab + 1 = 0 \) (i)
2. \( b - a^2 = 0 \) (ii)

From (ii), we have:
\[ b = a^2 \]

### Step 6: Substitute \( b \) into equation (i)
Substituting \( b = a^2 \) into (i):
\[ -a^3 + 1 = 0 \]
This simplifies to:
\[ a^3 = 1 \implies a = 1 \]

### Step 7: Find \( b \)
Substituting \( a = 1 \) back into \( b = a^2 \):
\[ b = 1^2 = 1 \]

### Step 8: Calculate \( 2a^2 + b \)
Now, we can calculate \( 2a^2 + b \):
\[ 2a^2 + b = 2(1^2) + 1 = 2 + 1 = 3 \]

### Final Answer
Thus, the value of \( 2a^2 + b \) is \( \boxed{3} \).