Algebra Examples

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

Step 1

Complete the square for $x^{2} + x$ .

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Step 1.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 1$

$c = 0$

Step 1.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{1}{2 \cdot 1}$

Step 1.3.2

Cancel the common factor of $1$ .

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Step 1.3.2.1

Cancel the common factor.

$d = \frac{1}{2 \cdot 1}$

Step 1.3.2.2

Rewrite the expression.

$d = \frac{1}{2}$

Step 1.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{1^{2}}{4 \cdot 1}$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

One to any power is one.

$e = 0 - \frac{1}{4 \cdot 1}$

Step 1.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{1}{4}$

Step 1.4.2.2

Subtract $\frac{1}{4}$ from $0$ .

$e = - \frac{1}{4}$

Step 1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ .

${(x + \frac{1}{2})}^{2} - \frac{1}{4}$

Step 2

Substitute ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ for $x^{2} + x$ in the equation $x^{2} + y^{2} + x = 0$ .

${(x + \frac{1}{2})}^{2} - \frac{1}{4} + y^{2} = 0$

Step 3

Move $- \frac{1}{4}$ to the right side of the equation by adding $\frac{1}{4}$ to both sides.

${(x + \frac{1}{2})}^{2} + y^{2} = 0 + \frac{1}{4}$

Step 4

Add $0$ and $\frac{1}{4}$ .

${(x + \frac{1}{2})}^{2} + y^{2} = \frac{1}{4}$