Solve for x (x-3)^2+(x-6)^2=x^2

${(x - 3)}^{2} + {(x - 6)}^{2} = x^{2}$

Step 1

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

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

${(x - 3)}^{2} + {(x - 6)}^{2} - x^{2} = 0$

Simplify each term.

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

$(x - 3) (x - 3) + {(x - 6)}^{2} - x^{2} = 0$

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

Apply the distributive property.

$x (x - 3) - 3 (x - 3) + {(x - 6)}^{2} - x^{2} = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 3 - 3 (x - 3) + {(x - 6)}^{2} - x^{2} = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3 + {(x - 6)}^{2} - x^{2} = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3 + {(x - 6)}^{2} - x^{2} = 0$

Move $- 3$ to the left of $x$ .

$x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3 + {(x - 6)}^{2} - x^{2} = 0$

Multiply $- 3$ by $- 3$ .

$x^{2} - 3 x - 3 x + 9 + {(x - 6)}^{2} - x^{2} = 0$

Subtract $3 x$ from $- 3 x$ .

$x^{2} - 6 x + 9 + {(x - 6)}^{2} - x^{2} = 0$

Rewrite ${(x - 6)}^{2}$ as $(x - 6) (x - 6)$ .

$x^{2} - 6 x + 9 + (x - 6) (x - 6) - x^{2} = 0$

Expand $(x - 6) (x - 6)$ using the FOIL Method.

Apply the distributive property.

$x^{2} - 6 x + 9 + x (x - 6) - 6 (x - 6) - x^{2} = 0$

Apply the distributive property.

$x^{2} - 6 x + 9 + x \cdot x + x \cdot - 6 - 6 (x - 6) - x^{2} = 0$

Apply the distributive property.

$x^{2} - 6 x + 9 + x \cdot x + x \cdot - 6 - 6 x - 6 \cdot - 6 - x^{2} = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} - 6 x + 9 + x^{2} + x \cdot - 6 - 6 x - 6 \cdot - 6 - x^{2} = 0$

Move $- 6$ to the left of $x$ .

$x^{2} - 6 x + 9 + x^{2} - 6 \cdot x - 6 x - 6 \cdot - 6 - x^{2} = 0$

Multiply $- 6$ by $- 6$ .

$x^{2} - 6 x + 9 + x^{2} - 6 x - 6 x + 36 - x^{2} = 0$

Subtract $6 x$ from $- 6 x$ .

$x^{2} - 6 x + 9 + x^{2} - 12 x + 36 - x^{2} = 0$

Combine the opposite terms in $x^{2} - 6 x + 9 + x^{2} - 12 x + 36 - x^{2}$ .

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

$- 6 x + 9 + x^{2} - 12 x + 36 + 0 = 0$

Add $- 6 x + 9 + x^{2} - 12 x + 36$ and $0$ .

$- 6 x + 9 + x^{2} - 12 x + 36 = 0$

Subtract $12 x$ from $- 6 x$ .

$- 18 x + 9 + x^{2} + 36 = 0$

Add $9$ and $36$ .

$- 18 x + x^{2} + 45 = 0$

Step 2

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$- 18 u + u^{2} + 45 = 0$

Factor $- 18 u + u^{2} + 45$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $45$ and whose sum is $- 18$ .

$- 15, - 3$

Write the factored form using these integers.

$(u - 15) (u - 3) = 0$

Replace all occurrences of $u$ with $x$ .

$(x - 15) (x - 3) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 15 = 0$

$x - 3 = 0$

Step 4

Set $x - 15$ equal to $0$ and solve for $x$ .

Set $x - 15$ equal to $0$ .

$x - 15 = 0$

Add $15$ to both sides of the equation.

$x = 15$

Step 5

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 6

The final solution is all the values that make $(x - 15) (x - 3) = 0$ true.

$x = 15, 3$

Solve for x (x-3)^2+(x-6)^2=x^2

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