Calculus Examples

2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9

$2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9$

Step 1

Complete the square for

2 x^{2} + x

$2 x^{2} + x$ .

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Step 1.1

Use the form

a x^{2} + b x + c

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

a

$a$ ,

b

$b$ , and

c

$c$ .

a = 2

$a = 2$

b = 1

$b = 1$

c = 0

$c = 0$

Step 1.2

Consider the vertex form of a parabola.

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

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

Step 1.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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Step 1.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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

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

Step 1.3.2

Multiply

2

$2$ by

2

$2$ .

d = \frac{1}{4}

$d = \frac{1}{4}$

d = \frac{1}{4}

$d = \frac{1}{4}$

Step 1.4

Find the value of

e

$e$ using the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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Step 1.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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

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

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 2}

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

Step 1.4.2.1.2

Multiply

4

$4$ by

2

$2$ .

e = 0 - \frac{1}{8}

$e = 0 - \frac{1}{8}$

e = 0 - \frac{1}{8}

$e = 0 - \frac{1}{8}$

Step 1.4.2.2

Subtract

\frac{1}{8}

$\frac{1}{8}$ from

0

$0$ .

e = - \frac{1}{8}

$e = - \frac{1}{8}$

e = - \frac{1}{8}

$e = - \frac{1}{8}$

e = - \frac{1}{8}

$e = - \frac{1}{8}$

Step 1.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}$ .

2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}$

2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}$

Step 2

Substitute

2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} - \frac{1}{8}$ for

2 x^{2} + x

$2 x^{2} + x$ in the equation

2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9

$2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9$ .

2 {(x + \frac{1}{4})}^{2} - \frac{1}{8} + 2 y^{2} + 2 z^{2} + y + z = 9

$2 {(x + \frac{1}{4})}^{2} - \frac{1}{8} + 2 y^{2} + 2 z^{2} + y + z = 9$

Step 3

Move

- \frac{1}{8}

$- \frac{1}{8}$ to the right side of the equation by adding

\frac{1}{8}

$\frac{1}{8}$ to both sides.

2 {(x + \frac{1}{4})}^{2} + 2 y^{2} + 2 z^{2} + y + z = 9 + \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} + 2 y^{2} + 2 z^{2} + y + z = 9 + \frac{1}{8}$

Step 4

Complete the square for

2 y^{2} + y

$2 y^{2} + y$ .

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Step 4.1

Use the form

a x^{2} + b x + c

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

a

$a$ ,

b

$b$ , and

c

$c$ .

a = 2

$a = 2$

b = 1

$b = 1$

c = 0

$c = 0$

Step 4.2

Consider the vertex form of a parabola.

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

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

Step 4.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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Step 4.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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

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

Step 4.3.2

Multiply

2

$2$ by

2

$2$ .

d = \frac{1}{4}

$d = \frac{1}{4}$

d = \frac{1}{4}

$d = \frac{1}{4}$

Step 4.4

Find the value of

e

$e$ using the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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Step 4.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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

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

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

One to any power is one.

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

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

Step 4.4.2.1.2

Multiply

4

$4$ by

2

$2$ .

e = 0 - \frac{1}{8}

$e = 0 - \frac{1}{8}$

e = 0 - \frac{1}{8}

$e = 0 - \frac{1}{8}$

Step 4.4.2.2

Subtract

\frac{1}{8}

$\frac{1}{8}$ from

0

$0$ .

e = - \frac{1}{8}

$e = - \frac{1}{8}$

e = - \frac{1}{8}

$e = - \frac{1}{8}$

e = - \frac{1}{8}

$e = - \frac{1}{8}$

Step 4.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}$ .

2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}$

2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}$

Step 5

Substitute

2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(y + \frac{1}{4})}^{2} - \frac{1}{8}$ for

2 y^{2} + y

$2 y^{2} + y$ in the equation

2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9

$2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9$ .

2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} - \frac{1}{8} = 9 + \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} - \frac{1}{8} = 9 + \frac{1}{8}$

Step 6

Move

- \frac{1}{8}

$- \frac{1}{8}$ to the right side of the equation by adding

\frac{1}{8}

$\frac{1}{8}$ to both sides.

2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} = 9 + \frac{1}{8} + \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} = 9 + \frac{1}{8} + \frac{1}{8}$

Step 7

Complete the square for

2 z^{2} + z

$2 z^{2} + z$ .

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Step 7.1

Use the form

a x^{2} + b x + c

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

a

$a$ ,

b

$b$ , and

c

$c$ .

a = 2

$a = 2$

b = 1

$b = 1$

c = 0

$c = 0$

Step 7.2

Consider the vertex form of a parabola.

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

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

Step 7.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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Step 7.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

$d = \frac{b}{2 a}$ .

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

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

Step 7.3.2

Multiply

2

$2$ by

2

$2$ .

d = \frac{1}{4}

$d = \frac{1}{4}$

d = \frac{1}{4}

$d = \frac{1}{4}$

Step 7.4

Find the value of

e

$e$ using the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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Step 7.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

e = c - \frac{b^{2}}{4 a}

$e = c - \frac{b^{2}}{4 a}$ .

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

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

Step 7.4.2

Simplify the right side.

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Step 7.4.2.1

Simplify each term.

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Step 7.4.2.1.1

One to any power is one.

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

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

Step 7.4.2.1.2

Multiply

4

$4$ by

2

$2$ .

e = 0 - \frac{1}{8}

$e = 0 - \frac{1}{8}$

e = 0 - \frac{1}{8}

$e = 0 - \frac{1}{8}$

Step 7.4.2.2

Subtract

\frac{1}{8}

$\frac{1}{8}$ from

0

$0$ .

e = - \frac{1}{8}

$e = - \frac{1}{8}$

e = - \frac{1}{8}

$e = - \frac{1}{8}$

e = - \frac{1}{8}

$e = - \frac{1}{8}$

Step 7.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}$ .

2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}$

2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}$

Step 8

Substitute

2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}

$2 {(z + \frac{1}{4})}^{2} - \frac{1}{8}$ for

2 z^{2} + z

$2 z^{2} + z$ in the equation

2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9

$2 x^{2} + 2 y^{2} + 2 z^{2} + x + y + z = 9$ .

2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} - \frac{1}{8} = 9 + \frac{1}{8} + \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} - \frac{1}{8} = 9 + \frac{1}{8} + \frac{1}{8}$

Step 9

Move

- \frac{1}{8}

$- \frac{1}{8}$ to the right side of the equation by adding

\frac{1}{8}

$\frac{1}{8}$ to both sides.

2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = 9 + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = 9 + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$

Step 10

Simplify

9 + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}

$9 + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$ .

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Step 10.1

Combine the numerators over the common denominator.

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = 9 + \frac{1 + 1 + 1}{8}$

Step 10.2

Simplify by adding numbers.

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Step 10.2.1

Add

$1$ and

$1$ .

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = 9 + \frac{2 + 1}{8}$

Step 10.2.2

Add

$2$ and

$1$ .

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = 9 + \frac{3}{8}$

Step 10.3

To write

$9$ as a fraction with a common denominator, multiply by

$\frac{8}{8}$ .

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = 9 \cdot \frac{8}{8} + \frac{3}{8}$

Step 10.4

Combine

$9$ and

$\frac{8}{8}$ .

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = \frac{9 \cdot 8}{8} + \frac{3}{8}$

Step 10.5

Combine the numerators over the common denominator.

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = \frac{9 \cdot 8 + 3}{8}$

Step 10.6

Simplify the numerator.

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Step 10.6.1

Multiply

$9$ by

$8$ .

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = \frac{72 + 3}{8}$

Step 10.6.2

Add

$72$ and

$3$ .

$2 {(x + \frac{1}{4})}^{2} + 2 {(y + \frac{1}{4})}^{2} + 2 {(z + \frac{1}{4})}^{2} = \frac{75}{8}$

Step 11