Solve for k square root of 2(x-k)=x-k

$\sqrt{2 (x - k)} = x - k$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{2 (x - k)}}^{2} = {(x - k)}^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 (x - k)}$ as ${(2 (x - k))}^{\frac{1}{2}}$ .

${({(2 (x - k))}^{\frac{1}{2}})}^{2} = {(x - k)}^{2}$

Simplify the left side.

Simplify ${({(2 (x - k))}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 (x - k))}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(2 (x - k))}^{\frac{1}{2} \cdot 2} = {(x - k)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 (x - k))}^{\frac{1}{2} \cdot 2} = {(x - k)}^{2}$

Rewrite the expression.

${(2 (x - k))}^{1} = {(x - k)}^{2}$

Apply the distributive property.

${(2 x + 2 (- k))}^{1} = {(x - k)}^{2}$

Multiply.

Multiply $- 1$ by $2$ .

${(2 x - 2 k)}^{1} = {(x - k)}^{2}$

Simplify.

$2 x - 2 k = {(x - k)}^{2}$

Simplify the right side.

Simplify ${(x - k)}^{2}$ .

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

$2 x - 2 k = (x - k) (x - k)$

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

Apply the distributive property.

$2 x - 2 k = x (x - k) - k (x - k)$

Apply the distributive property.

$2 x - 2 k = x \cdot x + x (- k) - k (x - k)$

Apply the distributive property.

$2 x - 2 k = x \cdot x + x (- k) - k x - k (- k)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$2 x - 2 k = x^{2} + x (- k) - k x - k (- k)$

Rewrite using the commutative property of multiplication.

$2 x - 2 k = x^{2} - x k - k x - k (- k)$

Rewrite using the commutative property of multiplication.

$2 x - 2 k = x^{2} - x k - k x - 1 \cdot - 1 k \cdot k$

Multiply $k$ by $k$ by adding the exponents.

Move $k$ .

$2 x - 2 k = x^{2} - x k - k x - 1 \cdot - 1 (k \cdot k)$

Multiply $k$ by $k$ .

$2 x - 2 k = x^{2} - x k - k x - 1 \cdot - 1 k^{2}$

Multiply $- 1$ by $- 1$ .

$2 x - 2 k = x^{2} - x k - k x + 1 k^{2}$

Multiply $k^{2}$ by $1$ .

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

Subtract $k x$ from $- x k$ .

Move $x$ .

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

Subtract $k x$ from $- k x$ .

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

Step 3

Solve for $k$ .

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

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

Add $2 k$ to both sides of the equation.

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

Subtract $2 x$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2 x + 2$ , and $c = x^{2} - 2 x$ into the quadratic formula and solve for $k$ .

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

Simplify.

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

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

Add parentheses.

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

Let $u = 1 \cdot (x^{2} - 2 x)$ . Substitute $u$ for all occurrences of $1 \cdot (x^{2} - 2 x)$ .

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

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

Expand $(- 2 x + 2) (- 2 x + 2)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 2$ by $- 2$ .

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

Multiply $2$ by $- 2$ .

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

Multiply $- 2$ by $2$ .

$k = \frac{2 x - 2 \pm \sqrt{4 x^{2} - 4 x - 4 x + 2 \cdot 2 - 4 \cdot u}}{2 \cdot 1}$

Multiply $2$ by $2$ .

$k = \frac{2 x - 2 \pm \sqrt{4 x^{2} - 4 x - 4 x + 4 - 4 \cdot u}}{2 \cdot 1}$

Subtract $4 x$ from $- 4 x$ .

$k = \frac{2 x - 2 \pm \sqrt{4 x^{2} - 8 x + 4 - 4 \cdot u}}{2 \cdot 1}$

$k = \frac{2 x - 2 \pm \sqrt{4 x^{2} - 8 x + 4 - 4 u}}{2 \cdot 1}$

Factor $4$ out of $4 x^{2} - 8 x + 4 - 4 u$ .

Factor $4$ out of $4 x^{2}$ .

$k = \frac{2 x - 2 \pm \sqrt{4 (x^{2}) - 8 x + 4 - 4 u}}{2 \cdot 1}$

Factor $4$ out of $- 8 x$ .

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

Factor $4$ out of $4$ .

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

Factor $4$ out of $- 4 u$ .

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

Factor $4$ out of $4 (x^{2}) + 4 (- 2 x)$ .

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

Factor $4$ out of $4 (x^{2} - 2 x) + 4 (1)$ .

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

Factor $4$ out of $4 (x^{2} - 2 x + 1) + 4 (- u)$ .

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

Replace all occurrences of $u$ with $1 \cdot (x^{2} - 2 x)$ .

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

Simplify.

Simplify each term.

Multiply $x^{2} - 2 x$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Combine the opposite terms in $x^{2} - 2 x + 1 - x^{2} + 2 x$ .

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

$k = \frac{2 x - 2 \pm \sqrt{4 (- 2 x + 1 + 0 + 2 x)}}{2 \cdot 1}$

Add $- 2 x + 1$ and $0$ .

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

Add $- 2 x$ and $2 x$ .

$k = \frac{2 x - 2 \pm \sqrt{4 (0 + 1)}}{2 \cdot 1}$

Add $0$ and $1$ .

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

Multiply $4$ by $1$ .

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

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

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

Pull terms out from under the radical, assuming positive real numbers.

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

Multiply $2$ by $1$ .

$k = \frac{2 x - 2 \pm 2}{2}$

The final answer is the combination of both solutions.

$k = x$

$k = x - 2$

$k = x$

$k = x - 2$

Solve for k square root of 2(x-k)=x-k

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