If `a+b=6, a-b=2` then what is the value of `2(a^(2)+b^(2))`?

Options

A. 20
B. 30
C. 40
D. 10

Answer

To find the value of \(2(a^2 + b^2)\) given the equations \(a + b = 6\) and \(a - b = 2\), we can follow these steps:

### Step 1: Square both equations
We start with the two equations:
1. \(a + b = 6\)
2. \(a - b = 2\)

First, we square both equations:

\[ (a + b)^2 = 6^2 \implies a^2 + 2ab + b^2 = 36 \quad \text{(Equation 1)} \]

\[ (a - b)^2 = 2^2 \implies a^2 - 2ab + b^2 = 4 \quad \text{(Equation 2)} \]

### Step 2: Set up the equations
From Equation 1, we have:
\[ a^2 + b^2 + 2ab = 36 \]

From Equation 2, we have:
\[ a^2 + b^2 - 2ab = 4 \]

### Step 3: Add the two equations
Now, we add Equation 1 and Equation 2:

\[ (a^2 + b^2 + 2ab) + (a^2 + b^2 - 2ab) = 36 + 4 \]

This simplifies to:

\[ 2a^2 + 2b^2 = 40 \]

### Step 4: Simplify the equation
Now, we can divide the entire equation by 2:

\[ a^2 + b^2 = 20 \]

### Step 5: Find \(2(a^2 + b^2)\)
Now, we need to find \(2(a^2 + b^2)\):

\[ 2(a^2 + b^2) = 2 \times 20 = 40 \]

### Final Answer
Thus, the value of \(2(a^2 + b^2)\) is \(40\).

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