Algebra Examples

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

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

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

Step 3.1.2

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 $b = 2 \cdot 1 y$ .

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

Step 3.1.3

Simplify.

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Step 3.1.3.1

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

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Step 3.1.3.1.1

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

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

Step 3.1.3.1.2

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

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

Step 3.1.3.1.3

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

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

Step 3.1.3.2

Add $- 1$ and $1$ .

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

Step 3.1.3.3

Combine exponents.

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Step 3.1.3.3.1

Multiply $0$ by $2$ .

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

Step 3.1.3.3.2

Multiply $0$ by $y$ .

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

Step 3.1.3.4

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

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Step 3.1.3.4.1

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

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

Step 3.1.3.4.2

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

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

Step 3.1.3.4.3

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

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

Step 3.1.3.5

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

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Step 3.1.3.5.1

Multiply $- 1$ by $2$ .

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

Step 3.1.3.5.2

Multiply $- 2$ by $1$ .

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

Step 3.1.3.6

Subtract $2$ from $- 2$ .

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

Step 3.1.3.7

Combine exponents.

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Step 3.1.3.7.1

Multiply $0$ by $y$ .

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

Step 3.1.3.7.2

Multiply $0$ by $- 4$ .

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

Step 3.1.4

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

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

Step 3.1.5

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

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

Step 3.1.6

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

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

Step 3.2

Multiply $2$ by $1$ .

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

Step 3.3

Cancel the common factor of $2$ .

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Step 3.3.1

Cancel the common factor.

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

Step 3.3.2

Divide $y$ by $1$ .

$x = y$

Step 4

The final answer is the combination of both solutions.

$x = y$ Double roots