Precalculus Examples

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$ ,

y = {(x - 3)}^{2}

$y = {(x - 3)}^{2}$

Step 1

Simplify

{(x - 3)}^{2}

${(x - 3)}^{2}$ .

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Step 1.1

Rewrite

{(x - 3)}^{2}

${(x - 3)}^{2}$ as

(x - 3) (x - 3)

$(x - 3) (x - 3)$ .

y = (x - 3) (x - 3)

$y = (x - 3) (x - 3)$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.2

Expand

(x - 3) (x - 3)

$(x - 3) (x - 3)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

y = x (x - 3) - 3 (x - 3)

$y = x (x - 3) - 3 (x - 3)$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.2.2

Apply the distributive property.

y = x \cdot x + x \cdot - 3 - 3 (x - 3)

$y = x \cdot x + x \cdot - 3 - 3 (x - 3)$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.2.3

Apply the distributive property.

y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3

$y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3

$y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply

x

$x$ by

x

$x$ .

y = x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3

$y = x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3.1.2

Move

- 3

$- 3$ to the left of

x

$x$ .

y = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3

$y = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3.1.3

Multiply

- 3

$- 3$ by

- 3

$- 3$ .

y = x^{2} - 3 x - 3 x + 9

$y = x^{2} - 3 x - 3 x + 9$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

y = x^{2} - 3 x - 3 x + 9

$y = x^{2} - 3 x - 3 x + 9$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3.2

Subtract

3 x

$3 x$ from

- 3 x

$- 3 x$ .

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 2

Replace all occurrences of

y

$y$ with

x^{2} - 6 x + 9

$x^{2} - 6 x + 9$ in each equation.

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Step 2.1

Replace all occurrences of

y

$y$ in

\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$ with

x^{2} - 6 x + 9

$x^{2} - 6 x + 9$ .

\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 9)}^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 9)}^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify

\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 9)}^{2}}{25}

$\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 9)}^{2}}{25}$ .

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Step 2.2.1.1

Simplify the numerator.

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Step 2.2.1.1.1

Factor using the perfect square rule.

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Step 2.2.1.1.1.1

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 3^{2})}^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 3^{2})}^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

6 x = 2 \cdot x \cdot 3

$6 x = 2 \cdot x \cdot 3$

Step 2.2.1.1.1.3

Rewrite the polynomial.

\frac{x^{2}}{9} + \frac{{(x^{2} - 2 \cdot x \cdot 3 + 3^{2})}^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(x^{2} - 2 \cdot x \cdot 3 + 3^{2})}^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.1.4

Factor using the perfect square trinomial rule

a^{2} - 2 a b + b^{2} = {(a - b)}^{2}

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where

a = x

$a = x$ and

b = 3

$b = 3$ .

\frac{x^{2}}{9} + \frac{{({(x - 3)}^{2})}^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{{({(x - 3)}^{2})}^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{{({(x - 3)}^{2})}^{2}}{25} = 1

$\frac{x^{2}}{9} + \frac{{({(x - 3)}^{2})}^{2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.2

Multiply the exponents in

{({(x - 3)}^{2})}^{2}

${({(x - 3)}^{2})}^{2}$ .

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Step 2.2.1.1.2.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

\frac{x^{2}}{9} + \frac{{(x - 3)}^{2 \cdot 2}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(x - 3)}^{2 \cdot 2}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.2.2

Multiply

2

$2$ by

2

$2$ .

\frac{x^{2}}{9} + \frac{{(x - 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(x - 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{{(x - 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(x - 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.3

Use the Binomial Theorem.

\frac{x^{2}}{9} + \frac{x^{4} + 4 x^{3} \cdot - 3 + 6 x^{2} {(- 3)}^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} + 4 x^{3} \cdot - 3 + 6 x^{2} {(- 3)}^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4

Simplify each term.

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Step 2.2.1.1.4.1

Multiply

- 3

$- 3$ by

4

$4$ .

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 6 x^{2} {(- 3)}^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 6 x^{2} {(- 3)}^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.2

Raise

- 3

$- 3$ to the power of

2

$2$ .

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 6 x^{2} \cdot 9 + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 6 x^{2} \cdot 9 + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.3

Multiply

9

$9$ by

6

$6$ .

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.4

Raise

- 3

$- 3$ to the power of

3

$3$ .

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} + 4 x \cdot - 27 + {(- 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} + 4 x \cdot - 27 + {(- 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.5

Multiply

- 27

$- 27$ by

4

$4$ .

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + {(- 3)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + {(- 3)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.6

Raise

- 3

$- 3$ to the power of

4

$4$ .

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.5

Make each term match the terms from the binomial theorem formula.

\frac{x^{2}}{9} + \frac{x^{4} - 4 \cdot (3 x^{3}) + 6 \cdot (3^{2} x^{2}) - 4 \cdot (3^{3} x) + 3^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{x^{4} - 4 \cdot (3 x^{3}) + 6 \cdot (3^{2} x^{2}) - 4 \cdot (3^{3} x) + 3^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.6

Factor

x^{4} - 4 \cdot 3 x^{3} + 6 \cdot 3^{2} x^{2} - 4 \cdot 3^{3} x + 3^{4}

$x^{4} - 4 \cdot 3 x^{3} + 6 \cdot 3^{2} x^{2} - 4 \cdot 3^{3} x + 3^{4}$ using the binomial theorem.

\frac{x^{2}}{9} + \frac{{(3 - x)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(3 - x)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2}}{9} + \frac{{(3 - x)}^{4}}{25} = 1

$\frac{x^{2}}{9} + \frac{{(3 - x)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.2

To write

\frac{x^{2}}{9}

$\frac{x^{2}}{9}$ as a fraction with a common denominator, multiply by

\frac{25}{25}

$\frac{25}{25}$ .

\frac{x^{2}}{9} \cdot \frac{25}{25} + \frac{{(3 - x)}^{4}}{25} = 1

$\frac{x^{2}}{9} \cdot \frac{25}{25} + \frac{{(3 - x)}^{4}}{25} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.3

To write

\frac{{(3 - x)}^{4}}{25}

$\frac{{(3 - x)}^{4}}{25}$ as a fraction with a common denominator, multiply by

\frac{9}{9}

$\frac{9}{9}$ .

\frac{x^{2}}{9} \cdot \frac{25}{25} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1

$\frac{x^{2}}{9} \cdot \frac{25}{25} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.4

Write each expression with a common denominator of

225

$225$ , by multiplying each by an appropriate factor of

1

$1$ .

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Step 2.2.1.4.1

Multiply

\frac{x^{2}}{9}

$\frac{x^{2}}{9}$ by

\frac{25}{25}

$\frac{25}{25}$ .

\frac{x^{2} \cdot 25}{9 \cdot 25} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1

$\frac{x^{2} \cdot 25}{9 \cdot 25} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.4.2

Multiply

9

$9$ by

25

$25$ .

\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.4.3

Multiply

\frac{{(3 - x)}^{4}}{25}

$\frac{{(3 - x)}^{4}}{25}$ by

\frac{9}{9}

$\frac{9}{9}$ .

\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{25 \cdot 9} = 1

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{25 \cdot 9} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.4.4

Multiply

25

$25$ by

9

$9$ .

\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{225} = 1

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{225} = 1

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.5

Combine the numerators over the common denominator.

\frac{x^{2} \cdot 25 + {(3 - x)}^{4} \cdot 9}{225} = 1

$\frac{x^{2} \cdot 25 + {(3 - x)}^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6

Simplify the numerator.

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Step 2.2.1.6.1

Move

25

$25$ to the left of

x^{2}

$x^{2}$ .

\frac{25 \cdot x^{2} + {(3 - x)}^{4} \cdot 9}{225} = 1

$\frac{25 \cdot x^{2} + {(3 - x)}^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.2

Use the Binomial Theorem.

\frac{25 x^{2} + (3^{4} + 4 \cdot (3^{3} (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (3^{4} + 4 \cdot (3^{3} (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3

Simplify each term.

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Step 2.2.1.6.3.1

Raise

3

$3$ to the power of

4

$4$ .

\frac{25 x^{2} + (81 + 4 \cdot (3^{3} (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 + 4 \cdot (3^{3} (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.2

Raise

3

$3$ to the power of

3

$3$ .

\frac{25 x^{2} + (81 + 4 \cdot (27 (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 + 4 \cdot (27 (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.3

Multiply

4

$4$ by

27

$27$ .

\frac{25 x^{2} + (81 + 108 (- x) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 + 108 (- x) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.4

Multiply

- 1

$- 1$ by

108

$108$ .

\frac{25 x^{2} + (81 - 108 x + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.5

Raise

3

$3$ to the power of

2

$2$ .

\frac{25 x^{2} + (81 - 108 x + 6 \cdot (9 {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 6 \cdot (9 {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.6

Multiply

6

$6$ by

9

$9$ .

\frac{25 x^{2} + (81 - 108 x + 54 {(- x)}^{2} + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 {(- x)}^{2} + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.7

Apply the product rule to

- x

$- x$ .

\frac{25 x^{2} + (81 - 108 x + 54 ({(- 1)}^{2} x^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 ({(- 1)}^{2} x^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.8

Raise

- 1

$- 1$ to the power of

2

$2$ .

\frac{25 x^{2} + (81 - 108 x + 54 (1 x^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 (1 x^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.9

Multiply

x^{2}

$x^{2}$ by

1

$1$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.10

Multiply

4

$4$ by

3

$3$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 {(- x)}^{3} + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 {(- x)}^{3} + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.11

Apply the product rule to

- x

$- x$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 ({(- 1)}^{3} x^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 ({(- 1)}^{3} x^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.12

Raise

- 1

$- 1$ to the power of

3

$3$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 (- x^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 (- x^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.13

Multiply

- 1

$- 1$ by

12

$12$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + {(- x)}^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + {(- x)}^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.14

Apply the product rule to

- x

$- x$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + {(- 1)}^{4} x^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + {(- 1)}^{4} x^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.15

Raise

- 1

$- 1$ to the power of

4

$4$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + 1 x^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + 1 x^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.16

Multiply

x^{4}

$x^{4}$ by

1

$1$ .

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}) \cdot 9}{225} = 1

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}) \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.4

Apply the distributive property.

\frac{25 x^{2} + 81 \cdot 9 - 108 x \cdot 9 + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1

$\frac{25 x^{2} + 81 \cdot 9 - 108 x \cdot 9 + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5

Simplify.

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Step 2.2.1.6.5.1

Multiply

81

$81$ by

9

$9$ .

\frac{25 x^{2} + 729 - 108 x \cdot 9 + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1

$\frac{25 x^{2} + 729 - 108 x \cdot 9 + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.2

Multiply

9

$9$ by

- 108

$- 108$ .

\frac{25 x^{2} + 729 - 972 x + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1

$\frac{25 x^{2} + 729 - 972 x + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.3

Multiply

9

$9$ by

54

$54$ .

\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.4

Multiply

9

$9$ by

- 12

$- 12$ .

\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + x^{4} \cdot 9}{225} = 1

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + x^{4} \cdot 9}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.5

Move

9

$9$ to the left of

x^{4}

$x^{4}$ .

\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + 9 \cdot x^{4}}{225} = 1

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + 9 \cdot x^{4}}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + 9 \cdot x^{4}}{225} = 1

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + 9 \cdot x^{4}}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.6

Add

25 x^{2}

$25 x^{2}$ and

486 x^{2}

$486 x^{2}$ .

\frac{511 x^{2} + 729 - 972 x - 108 x^{3} + 9 x^{4}}{225} = 1

$\frac{511 x^{2} + 729 - 972 x - 108 x^{3} + 9 x^{4}}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.7

Reorder terms.

\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 3

Graph each side of the equation. The solution is the x-value of the point of intersection.

x \approx 0.80535522, 3

$x \approx 0.80535522, 3$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 4

Remove any equations from the system that are always true.

x \approx 0.80535522

$x \approx 0.80535522$

y = x^{2} - 6 x + 9

$y = x^{2} - 6 x + 9$

Step 5