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Sagot :
Let’s solve the equation [tex]\(x^2 + 2xy + y^2 = 8\)[/tex] for one variable in terms of the other.
### Step 1: Rewriting the Equation
The given equation is:
[tex]\[ x^2 + 2xy + y^2 = 8 \][/tex]
### Step 2: Solving for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]
We will rewrite the equation to a standard form to solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 + 2xy + y^2 - 8 = 0 \][/tex]
This is a quadratic equation in [tex]\(x\)[/tex]:
[tex]\[ x^2 + 2xy + (y^2 - 8) = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2y\)[/tex], and [tex]\(c = y^2 - 8\)[/tex].
Applying the quadratic formula:
[tex]\[ x = \frac{-2y \pm \sqrt{(2y)^2 - 4 \cdot 1 \cdot (y^2 - 8)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2y \pm \sqrt{4y^2 - 4(y^2 - 8)}}{2} \][/tex]
[tex]\[ x = \frac{-2y \pm \sqrt{4y^2 - 4y^2 + 32}}{2} \][/tex]
[tex]\[ x = \frac{-2y \pm \sqrt{32}}{2} \][/tex]
[tex]\[ x = \frac{-2y \pm 4\sqrt{2}}{2} \][/tex]
[tex]\[ x = -y \pm 2\sqrt{2} \][/tex]
So, the solutions for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex] are:
[tex]\[ x = -y - 2\sqrt{2} \quad \text{and} \quad x = -y + 2\sqrt{2} \][/tex]
### Step 3: Solving for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]
Next, we rewrite the equation to solve for [tex]\(y\)[/tex]:
[tex]\[ y^2 + 2xy + x^2 - 8 = 0 \][/tex]
This is a quadratic equation in [tex]\(y\)[/tex]:
[tex]\[ y^2 + 2xy + (x^2 - 8) = 0 \][/tex]
Again, applying the quadratic formula for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot (x^2 - 8)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-2x \pm \sqrt{4x^2 - 4(x^2 - 8)}}{2} \][/tex]
[tex]\[ y = \frac{-2x \pm \sqrt{4x^2 - 4x^2 + 32}}{2} \][/tex]
[tex]\[ y = \frac{-2x \pm \sqrt{32}}{2} \][/tex]
[tex]\[ y = \frac{-2x \pm 4\sqrt{2}}{2} \][/tex]
[tex]\[ y = -x \pm 2\sqrt{2} \][/tex]
So, the solutions for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex] are:
[tex]\[ y = -x - 2\sqrt{2} \quad \text{and} \quad y = -x + 2\sqrt{2} \][/tex]
### Conclusion
The solutions to the equation [tex]\( x^2 + 2xy + y^2 = 8 \)[/tex] are:
1. [tex]\( x = -y - 2\sqrt{2} \)[/tex]
2. [tex]\( x = -y + 2\sqrt{2} \)[/tex]
And equivalently:
1. [tex]\( y = -x - 2\sqrt{2} \)[/tex]
2. [tex]\( y = -x + 2\sqrt{2} \)[/tex]
### Step 1: Rewriting the Equation
The given equation is:
[tex]\[ x^2 + 2xy + y^2 = 8 \][/tex]
### Step 2: Solving for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]
We will rewrite the equation to a standard form to solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 + 2xy + y^2 - 8 = 0 \][/tex]
This is a quadratic equation in [tex]\(x\)[/tex]:
[tex]\[ x^2 + 2xy + (y^2 - 8) = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2y\)[/tex], and [tex]\(c = y^2 - 8\)[/tex].
Applying the quadratic formula:
[tex]\[ x = \frac{-2y \pm \sqrt{(2y)^2 - 4 \cdot 1 \cdot (y^2 - 8)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2y \pm \sqrt{4y^2 - 4(y^2 - 8)}}{2} \][/tex]
[tex]\[ x = \frac{-2y \pm \sqrt{4y^2 - 4y^2 + 32}}{2} \][/tex]
[tex]\[ x = \frac{-2y \pm \sqrt{32}}{2} \][/tex]
[tex]\[ x = \frac{-2y \pm 4\sqrt{2}}{2} \][/tex]
[tex]\[ x = -y \pm 2\sqrt{2} \][/tex]
So, the solutions for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex] are:
[tex]\[ x = -y - 2\sqrt{2} \quad \text{and} \quad x = -y + 2\sqrt{2} \][/tex]
### Step 3: Solving for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]
Next, we rewrite the equation to solve for [tex]\(y\)[/tex]:
[tex]\[ y^2 + 2xy + x^2 - 8 = 0 \][/tex]
This is a quadratic equation in [tex]\(y\)[/tex]:
[tex]\[ y^2 + 2xy + (x^2 - 8) = 0 \][/tex]
Again, applying the quadratic formula for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot (x^2 - 8)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-2x \pm \sqrt{4x^2 - 4(x^2 - 8)}}{2} \][/tex]
[tex]\[ y = \frac{-2x \pm \sqrt{4x^2 - 4x^2 + 32}}{2} \][/tex]
[tex]\[ y = \frac{-2x \pm \sqrt{32}}{2} \][/tex]
[tex]\[ y = \frac{-2x \pm 4\sqrt{2}}{2} \][/tex]
[tex]\[ y = -x \pm 2\sqrt{2} \][/tex]
So, the solutions for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex] are:
[tex]\[ y = -x - 2\sqrt{2} \quad \text{and} \quad y = -x + 2\sqrt{2} \][/tex]
### Conclusion
The solutions to the equation [tex]\( x^2 + 2xy + y^2 = 8 \)[/tex] are:
1. [tex]\( x = -y - 2\sqrt{2} \)[/tex]
2. [tex]\( x = -y + 2\sqrt{2} \)[/tex]
And equivalently:
1. [tex]\( y = -x - 2\sqrt{2} \)[/tex]
2. [tex]\( y = -x + 2\sqrt{2} \)[/tex]
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