Answer :
Certainly! Let's take a detailed look at the equation:
[tex]\[ \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 = 0 \][/tex]
### Step-by-step Solution
1. Identify and Rewrite the Equation:
The given equation is:
[tex]\[ \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 \][/tex]
2. Factor the Equation:
We need to factorize the quadratic expression [tex]\( \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 \)[/tex].
By examining the expression, it can be rewritten and factored as follows:
[tex]\[ 2 \left( 0.25 x^2 + 1.0 xy + 0.25 y^2 \right) \][/tex]
3. Using the Factored Form:
The factored form of the equation is therefore:
[tex]\[ 2 \left( 0.25 x^2 + 1.0 xy + 0.25 y^2 \right) \][/tex]
By having the factored equation, you can now see the form of the expression clearly.
### Simplified Form:
To solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] such that this equation equals zero, we set the factored expression to zero:
[tex]\[ 2 \left( 0.25 x^2 + 1.0 xy + 0.25 y^2 \right) = 0 \][/tex]
Since [tex]\( 2 \)[/tex] is a constant and cannot be zero, we focus on the expression inside the parentheses:
[tex]\[ 0.25 x^2 + 1.0 xy + 0.25 y^2 = 0 \][/tex]
4. Solving the Simplified Expression:
Notice that the simplified expression [tex]\( 0.25 x^2 + 1.0 xy + 0.25 y^2 \)[/tex] resembles the expanded form of a perfect square trinomial.
Rewriting it:
[tex]\[ 0.25 (x^2 + 4xy + y^2) \][/tex]
Factoring the perfect square:
[tex]\[ 0.25 (x + y)^2 = 0 \][/tex]
5. Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
We now solve:
[tex]\[ (x + y)^2 = 0 \][/tex]
Taking the square root on both sides:
[tex]\[ x + y = 0 \][/tex]
6. Conclusion:
Therefore, the solution to the equation is:
[tex]\[ x + y = 0 \quad \Rightarrow \quad x = -y \][/tex]
This tells us that [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are opposites of each other when the equation [tex]\( \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 = 0 \)[/tex] holds true.
[tex]\[ \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 = 0 \][/tex]
### Step-by-step Solution
1. Identify and Rewrite the Equation:
The given equation is:
[tex]\[ \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 \][/tex]
2. Factor the Equation:
We need to factorize the quadratic expression [tex]\( \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 \)[/tex].
By examining the expression, it can be rewritten and factored as follows:
[tex]\[ 2 \left( 0.25 x^2 + 1.0 xy + 0.25 y^2 \right) \][/tex]
3. Using the Factored Form:
The factored form of the equation is therefore:
[tex]\[ 2 \left( 0.25 x^2 + 1.0 xy + 0.25 y^2 \right) \][/tex]
By having the factored equation, you can now see the form of the expression clearly.
### Simplified Form:
To solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] such that this equation equals zero, we set the factored expression to zero:
[tex]\[ 2 \left( 0.25 x^2 + 1.0 xy + 0.25 y^2 \right) = 0 \][/tex]
Since [tex]\( 2 \)[/tex] is a constant and cannot be zero, we focus on the expression inside the parentheses:
[tex]\[ 0.25 x^2 + 1.0 xy + 0.25 y^2 = 0 \][/tex]
4. Solving the Simplified Expression:
Notice that the simplified expression [tex]\( 0.25 x^2 + 1.0 xy + 0.25 y^2 \)[/tex] resembles the expanded form of a perfect square trinomial.
Rewriting it:
[tex]\[ 0.25 (x^2 + 4xy + y^2) \][/tex]
Factoring the perfect square:
[tex]\[ 0.25 (x + y)^2 = 0 \][/tex]
5. Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
We now solve:
[tex]\[ (x + y)^2 = 0 \][/tex]
Taking the square root on both sides:
[tex]\[ x + y = 0 \][/tex]
6. Conclusion:
Therefore, the solution to the equation is:
[tex]\[ x + y = 0 \quad \Rightarrow \quad x = -y \][/tex]
This tells us that [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are opposites of each other when the equation [tex]\( \frac{1}{2} x^2 + 2xy + \frac{1}{2} y^2 = 0 \)[/tex] holds true.