Algebra Examples

$x^{2} = 2 x + x (2 x - 4) + y$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 x + x (2 x - 4) + y = x^{2}$

Step 2

Simplify $2 x + x (2 x - 4) + y$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$2 x + x (2 x) + x \cdot - 4 + y = x^{2}$

Step 2.1.2

Rewrite using the commutative property of multiplication.

$2 x + 2 x \cdot x + x \cdot - 4 + y = x^{2}$

Step 2.1.3

Move $- 4$ to the left of $x$ .

$2 x + 2 x \cdot x - 4 \cdot x + y = x^{2}$

Step 2.1.4

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.4.1

Move $x$ .

$2 x + 2 (x \cdot x) - 4 \cdot x + y = x^{2}$

Step 2.1.4.2

Multiply $x$ by $x$ .

$2 x + 2 x^{2} - 4 \cdot x + y = x^{2}$

$2 x + 2 x^{2} - 4 x + y = x^{2}$

Step 2.2

Subtract $4 x$ from $2 x$ .

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

Step 3

Move all terms containing $x$ to the left side of the equation.

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Step 3.1

Subtract $x^{2}$ from both sides of the equation.

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

Step 3.2

Subtract $x^{2}$ from $2 x^{2}$ .

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

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

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

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

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 2$ to the power of $2$ .

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

Step 6.1.2

Multiply $- 4$ by $1$ .

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

Step 6.1.3

Factor $4$ out of $4 - 4 y$ .

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Step 6.1.3.1

Factor $4$ out of $4$ .

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

Step 6.1.3.2

Factor $4$ out of $- 4 y$ .

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

Step 6.1.3.3

Factor $4$ out of $4 (1) + 4 (- y)$ .

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

Step 6.1.4

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

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Step 6.1.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 6.1.4.2

Rewrite $1$ as $1^{2}$ .

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

Step 6.1.5

Pull terms out from under the radical.

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

Step 6.1.6

One to any power is one.

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

Step 6.2

Multiply $2$ by $1$ .

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

Step 6.3

Simplify $\frac{2 \pm 2 \sqrt{1 - y}}{2}$ .

$x = 1 \pm \sqrt{1 - y}$

Step 7

The final answer is the combination of both solutions.

$x = 1 + \sqrt{1 - y}$

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