Solve for k square root of 5(x-k)=x-k

$\sqrt{5 (x - k)} = x - k$

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

${\sqrt{5 (x - k)}}^{2} = {(x - k)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 (x - k)}$ as ${(5 (x - k))}^{\frac{1}{2}}$ .

${({(5 (x - k))}^{\frac{1}{2}})}^{2} = {(x - k)}^{2}$

Simplify the left side.

Simplify ${({(5 (x - k))}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(5 (x - k))}^{\frac{1}{2}})}^{2}$ .

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

${(5 (x - k))}^{\frac{1}{2} \cdot 2} = {(x - k)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(5 (x - k))}^{\frac{1}{2} \cdot 2} = {(x - k)}^{2}$

Rewrite the expression.

${(5 (x - k))}^{1} = {(x - k)}^{2}$

Apply the distributive property.

${(5 x + 5 (- k))}^{1} = {(x - k)}^{2}$

Multiply.

Multiply $- 1$ by $5$ .

${(5 x - 5 k)}^{1} = {(x - k)}^{2}$

Simplify.

$5 x - 5 k = {(x - k)}^{2}$

Simplify the right side.

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

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

$5 x - 5 k = (x - k) (x - k)$

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

Apply the distributive property.

$5 x - 5 k = x (x - k) - k (x - k)$

Apply the distributive property.

$5 x - 5 k = x \cdot x + x (- k) - k (x - k)$

Apply the distributive property.

$5 x - 5 k = x \cdot x + x (- k) - k x - k (- k)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$5 x - 5 k = x^{2} + x (- k) - k x - k (- k)$

Rewrite using the commutative property of multiplication.

$5 x - 5 k = x^{2} - x k - k x - k (- k)$

Rewrite using the commutative property of multiplication.

$5 x - 5 k = x^{2} - x k - k x - 1 \cdot - 1 k \cdot k$

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

Move $k$ .

$5 x - 5 k = x^{2} - x k - k x - 1 \cdot - 1 (k \cdot k)$

Multiply $k$ by $k$ .

$5 x - 5 k = x^{2} - x k - k x - 1 \cdot - 1 k^{2}$

Multiply $- 1$ by $- 1$ .

$5 x - 5 k = x^{2} - x k - k x + 1 k^{2}$

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

$5 x - 5 k = x^{2} - x k - k x + k^{2}$

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

Move $x$ .

$5 x - 5 k = x^{2} - k x - k x + k^{2}$

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

$5 x - 5 k = x^{2} - 2 k x + k^{2}$

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.

$x^{2} - 2 k x + k^{2} = 5 x - 5 k$

Add $5 k$ to both sides of the equation.

$x^{2} - 2 k x + k^{2} + 5 k = 5 x$

Subtract $5 x$ from both sides of the equation.

$x^{2} - 2 k x + k^{2} + 5 k - 5 x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2 x + 5$ , and $c = x^{2} - 5 x$ into the quadratic formula and solve for $k$ .

$\frac{- (- 2 x + 5) \pm \sqrt{{(- 2 x + 5)}^{2} - 4 \cdot (1 \cdot (x^{2} - 5 x))}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Apply the distributive property.

$k = \frac{- (- 2 x) - 1 \cdot 5 \pm \sqrt{{(- 2 x + 5)}^{2} - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $- 2$ by $- 1$ .

$k = \frac{2 x - 1 \cdot 5 \pm \sqrt{{(- 2 x + 5)}^{2} - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $- 1$ by $5$ .

$k = \frac{2 x - 5 \pm \sqrt{{(- 2 x + 5)}^{2} - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Rewrite ${(- 2 x + 5)}^{2}$ as $(- 2 x + 5) (- 2 x + 5)$ .

$k = \frac{2 x - 5 \pm \sqrt{(- 2 x + 5) (- 2 x + 5) - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Expand $(- 2 x + 5) (- 2 x + 5)$ using the FOIL Method.

Apply the distributive property.

$k = \frac{2 x - 5 \pm \sqrt{- 2 x (- 2 x + 5) + 5 (- 2 x + 5) - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Apply the distributive property.

$k = \frac{2 x - 5 \pm \sqrt{- 2 x (- 2 x) - 2 x \cdot 5 + 5 (- 2 x + 5) - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Apply the distributive property.

$k = \frac{2 x - 5 \pm \sqrt{- 2 x (- 2 x) - 2 x \cdot 5 + 5 (- 2 x) + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$k = \frac{2 x - 5 \pm \sqrt{- 2 \cdot (- 2 x \cdot x) - 2 x \cdot 5 + 5 (- 2 x) + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

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

Move $x$ .

$k = \frac{2 x - 5 \pm \sqrt{- 2 \cdot (- 2 (x \cdot x)) - 2 x \cdot 5 + 5 (- 2 x) + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $x$ by $x$ .

$k = \frac{2 x - 5 \pm \sqrt{- 2 \cdot (- 2 x^{2}) - 2 x \cdot 5 + 5 (- 2 x) + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $- 2$ by $- 2$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 2 x \cdot 5 + 5 (- 2 x) + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $5$ by $- 2$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 10 x + 5 (- 2 x) + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $- 2$ by $5$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 10 x - 10 x + 5 \cdot 5 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $5$ by $5$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 10 x - 10 x + 25 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Subtract $10 x$ from $- 10 x$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 20 x + 25 - 4 \cdot 1 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 20 x + 25 - 4 \cdot (x^{2} - 5 x)}}{2 \cdot 1}$

Apply the distributive property.

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 20 x + 25 - 4 x^{2} - 4 (- 5 x)}}{2 \cdot 1}$

Multiply $- 5$ by $- 4$ .

$k = \frac{2 x - 5 \pm \sqrt{4 x^{2} - 20 x + 25 - 4 x^{2} + 20 x}}{2 \cdot 1}$

Subtract $4 x^{2}$ from $4 x^{2}$ .

$k = \frac{2 x - 5 \pm \sqrt{- 20 x + 25 + 0 + 20 x}}{2 \cdot 1}$

Add $- 20 x$ and $0$ .

$k = \frac{2 x - 5 \pm \sqrt{- 20 x + 25 + 20 x}}{2 \cdot 1}$

Add $- 20 x$ and $20 x$ .

$k = \frac{2 x - 5 \pm \sqrt{0 + 25}}{2 \cdot 1}$

Add $0$ and $25$ .

$k = \frac{2 x - 5 \pm \sqrt{25}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$k = \frac{2 x - 5 \pm \sqrt{5^{2}}}{2 \cdot 1}$

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

$k = \frac{2 x - 5 \pm 5}{2 \cdot 1}$

Multiply $2$ by $1$ .

$k = \frac{2 x - 5 \pm 5}{2}$

The final answer is the combination of both solutions.

$k = x$

$k = x - 5$

$k = x$

$k = x - 5$

Solve for k square root of 5(x-k)=x-k

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