IDNLearn.com connects you with a global community of knowledgeable individuals. Discover prompt and accurate answers from our community of experienced professionals.
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]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.
