Ass ignm ent (1) A binary operation $$ * $$ is degined 00 er R(the set of real numbers) by $$ x * y $$ $$ =x y+x^{2}+y^{2} $$ for all $$ x, y \in R $$ a) determine whether or not bed $$ H $$ is commutative b) If $$ x *(x+2)=49 $$, find $$ x $$

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Let's solve the problem step by step.

### Part (a): Determine if the operation $$ * $$ is commutative

The operation $$ * $$ is defined as:
$$
x * y = xy + x^2 + y^2
$$
To check if the operation is commutative, we need to verify if $$ x * y = y * x $$ for all $$ x, y \in \mathbb{R} $$.

1. Calculate $$ x * y $$:
   $$
   x * y = xy + x^2 + y^2
   $$

2. Calculate $$ y * x $$:
   $$
   y * x = yx + y^2 + x^2
   $$

3. Since multiplication of real numbers is commutative ($$ xy = yx $$), we can rewrite $$ y * x $$:
   $$
   y * x = xy + y^2 + x^2
   $$

4. Now, compare $$ x * y $$ and $$ y * x $$:
   $$
   x * y = xy + x^2 + y^2
   $$
   $$
   y * x = xy + y^2 + x^2
   $$

Since both expressions are equal, we conclude that the operation $$ * $$ is commutative.

### Part (b): Solve the equation $$ x * (x + 2) = 49 $$

Using the definition of the operation, we can substitute $$ y = x + 2 $$:
$$
x * (x + 2) = x(x + 2) + x^2 + (x + 2)^2
$$

Now, let's simplify the left-hand side:
1. Calculate $$ x(x + 2) $$:
   $$
   x(x + 2) = x^2 + 2x
   $$

2. Calculate $$ (x + 2)^2 $$:
   $$
   (x + 2)^2 = x^2 + 4x + 4
   $$

3. Combine all parts:
   $$
   x * (x + 2) = (x^2 + 2x) + x^2 + (x^2 + 4x + 4)
   $$
   $$
   = x^2 + 2x + x^2 + x^2 + 4x + 4
   $$
   $$
   = 3x^2 + 6x + 4
   $$

Now, set this equal to 49:
$$
3x^2 + 6x + 4 = 49
$$

4. Rearranging gives:
$$
3x^2 + 6x + 4 - 49 = 0
$$
$$
3x^2 + 6x - 45 = 0
$$

5. Divide the entire equation by 3:
$$
x^2 + 2x - 15 = 0
$$

Now, we can solve this quadratic equation using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$ a = 1, b = 2, c = -15 $$.

Calculating the discriminant:
$$
b^2 - 4ac = 2^2 - 4(1)(-15) = 4 + 60 = 64
$$

Now, substituting into the quadratic formula:
$$
x = \frac{-2 \pm \sqrt{64}}{2(1)} = \frac{-2 \pm 8}{2}
$$

Calculating the two possible values for $$ x $$:
1. $$ x = \frac{6}{2} = 3 $$
2. $$ x = \frac{-10}{2} = -5 $$

Thus, the solutions for $$ x $$ are:
$$
x = 3 \quad \text{or} \quad x = -5
$$

### Final Answers:
a) The operation $$ * $$ is commutative.  
b) The values of $$ x $$ are $$ 3 $$ and $$ -5 $$.
a) The operation $$ * $$ is commutative. b) $$ x = 3 $$ or $$ x = -5 $$.

Ass ignm ent (1) A binary operation \( * \) is degined 00 er R(the set of real numbers) by \( x * y \) \( =x y+x^{2}+y^{2} \) for all \( x, y \in R \) a) determine whether or not bed \( H \) is commutative b) If \( x *(x+2)=49 \), find \( x \)

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