Find the Product ((x^2+2xy+y^2)(x^2-2xy+y^2))/(2(x+y))=0

$\frac{(x^{2} + 2 x y + y^{2}) (x^{2} - 2 x y + y^{2})}{2 (x + y)} = 0$

Step 1

Set the numerator equal to zero.

$(x^{2} + 2 x y + y^{2}) (x^{2} - 2 x y + y^{2}) = 0$

Step 2

Solve the equation for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} + 2 x y + y^{2} = 0$

$x^{2} - 2 x y + y^{2} = 0$

Set $x^{2} + 2 x y + y^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2} + 2 x y + y^{2}$ equal to $0$ .

$x^{2} + 2 x y + y^{2} = 0$

Solve $x^{2} + 2 x y + y^{2} = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 2 y$ , and $c = y^{2}$ into the quadratic formula and solve for $x$ .

$\frac{- 2 y \pm \sqrt{{(2 y)}^{2} - 4 \cdot (1 \cdot y^{2})}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 1 \cdot y^{2}$ as ${(2 \cdot 1 y)}^{2}$ .

$x = \frac{- 2 y \pm \sqrt{{(2 y)}^{2} - {(2 \cdot (1 y))}^{2}}}{2 \cdot 1}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 y$ and $b = 2 \cdot 1 y$ .

$x = \frac{- 2 y \pm \sqrt{(2 y + 2 \cdot (1 y)) (2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Simplify.

Factor $2 y$ out of $2 y + 2 \cdot 1 y$ .

Factor $2 y$ out of $2 y$ .

$x = \frac{- 2 y \pm \sqrt{(2 y (1) + 2 \cdot (1 y)) (2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Factor $2 y$ out of $2 \cdot 1 y$ .

$x = \frac{- 2 y \pm \sqrt{(2 y (1) + 2 y (1)) (2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Factor $2 y$ out of $2 y (1) + 2 y (1)$ .

$x = \frac{- 2 y \pm \sqrt{2 y (1 + 1) (2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Add $1$ and $1$ .

$x = \frac{- 2 y \pm \sqrt{2 y \cdot (2 (2 y - (2 \cdot (1 y))))}}{2 \cdot 1}$

Multiply $2$ by $2$ .

$x = \frac{- 2 y \pm \sqrt{4 y (2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Combine exponents.

Multiply $2$ by $1$ .

$x = \frac{- 2 y \pm \sqrt{4 y (2 y - (2 y))}}{2 \cdot 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 2 y \pm \sqrt{4 y (2 y - 2 y)}}{2 \cdot 1}$

Subtract $2 y$ from $2 y$ .

$x = \frac{- 2 y \pm \sqrt{4 y \cdot 0}}{2 \cdot 1}$

Combine exponents.

Multiply $0$ by $4$ .

$x = \frac{- 2 y \pm \sqrt{0 y}}{2 \cdot 1}$

Multiply $0$ by $y$ .

$x = \frac{- 2 y \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{- 2 y \pm \sqrt{0^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 y \pm 0}{2 \cdot 1}$

$- 2 y$ plus or minus $0$ is $- 2 y$ .

$x = \frac{- 2 y}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 y}{2}$

Cancel the common factor of $- 2$ and $2$ .

Factor $2$ out of $- 2 y$ .

$x = \frac{2 (- y)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = \frac{2 (- y)}{2 (1)}$

Cancel the common factor.

$x = \frac{2 (- y)}{2 \cdot 1}$

Rewrite the expression.

$x = \frac{- y}{1}$

Divide $- y$ by $1$ .

$x = - y$

The final answer is the combination of both solutions.

$x = - y$ Double roots

Set $x^{2} - 2 x y + y^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2} - 2 x y + y^{2}$ equal to $0$ .

$x^{2} - 2 x y + y^{2} = 0$

Solve $x^{2} - 2 x y + y^{2} = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2 y$ , and $c = y^{2}$ into the quadratic formula and solve for $x$ .

$\frac{2 y \pm \sqrt{{(- 2 y)}^{2} - 4 \cdot (1 \cdot y^{2})}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 1 \cdot y^{2}$ as ${(2 \cdot 1 y)}^{2}$ .

$x = \frac{2 y \pm \sqrt{{(- 2 y)}^{2} - {(2 \cdot (1 y))}^{2}}}{2 \cdot 1}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 2 y$ and $b = 2 \cdot 1 y$ .

$x = \frac{2 y \pm \sqrt{(- 2 y + 2 \cdot (1 y)) (- 2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Simplify.

Factor $2 y$ out of $- 2 y + 2 \cdot 1 y$ .

Factor $2 y$ out of $- 2 y$ .

$x = \frac{2 y \pm \sqrt{(2 y (- 1) + 2 \cdot (1 y)) (- 2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Factor $2 y$ out of $2 \cdot 1 y$ .

$x = \frac{2 y \pm \sqrt{(2 y (- 1) + 2 y (1)) (- 2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Factor $2 y$ out of $2 y (- 1) + 2 y (1)$ .

$x = \frac{2 y \pm \sqrt{2 y (- 1 + 1) (- 2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Add $- 1$ and $1$ .

$x = \frac{2 y \pm \sqrt{2 y \cdot (0 (- 2 y - (2 \cdot (1 y))))}}{2 \cdot 1}$

Combine exponents.

Multiply $0$ by $2$ .

$x = \frac{2 y \pm \sqrt{0 y (- 2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Multiply $0$ by $y$ .

$x = \frac{2 y \pm \sqrt{0 (- 2 y - (2 \cdot (1 y)))}}{2 \cdot 1}$

Factor $y$ out of $- 2 y - 1 \cdot 2 \cdot 1 y$ .

Factor $y$ out of $- 2 y$ .

$x = \frac{2 y \pm \sqrt{0 (y \cdot - 2 - 1 \cdot 2 \cdot (1 y))}}{2 \cdot 1}$

Factor $y$ out of $- 1 \cdot 2 \cdot 1 y$ .

$x = \frac{2 y \pm \sqrt{0 (y \cdot - 2 + y (- 1 \cdot 2 \cdot 1))}}{2 \cdot 1}$

Factor $y$ out of $y \cdot - 2 + y (- 1 \cdot 2 \cdot 1)$ .

$x = \frac{2 y \pm \sqrt{0 (y (- 2 - 1 \cdot 2 \cdot 1))}}{2 \cdot 1}$

Multiply $- 1 \cdot 2 \cdot 1$ .

Multiply $- 1$ by $2$ .

$x = \frac{2 y \pm \sqrt{0 (y (- 2 - 2 \cdot 1))}}{2 \cdot 1}$

Multiply $- 2$ by $1$ .

$x = \frac{2 y \pm \sqrt{0 (y (- 2 - 2))}}{2 \cdot 1}$

Subtract $2$ from $- 2$ .

$x = \frac{2 y \pm \sqrt{0 y \cdot - 4}}{2 \cdot 1}$

Combine exponents.

Multiply $0$ by $y$ .

$x = \frac{2 y \pm \sqrt{0 \cdot - 4}}{2 \cdot 1}$

Multiply $0$ by $- 4$ .

$x = \frac{2 y \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{2 y \pm \sqrt{0^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 y \pm 0}{2 \cdot 1}$

$2 y$ plus or minus $0$ is $2 y$ .

$x = \frac{2 y}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 y}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \frac{2 y}{2}$

Divide $y$ by $1$ .

$x = y$

The final answer is the combination of both solutions.

$x = y$ Double roots

The final solution is all the values that make $(x^{2} + 2 x y + y^{2}) (x^{2} - 2 x y + y^{2}) = 0$ true.

$x = - y$

$x = y$

$x = - y$

$x = y$

Find the Product ((x^2+2xy+y^2)(x^2-2xy+y^2))/(2(x+y))=0

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