Solve for k k=1+ square root of 3k-5

$k = 1 + \sqrt{3 k - 5}$

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

$1 + \sqrt{3 k - 5} = k$

Subtract $1$ from both sides of the equation.

$\sqrt{3 k - 5} = k - 1$

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

${\sqrt{3 k - 5}}^{2} = {(k - 1)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 k - 5}$ as ${(3 k - 5)}^{\frac{1}{2}}$ .

${({(3 k - 5)}^{\frac{1}{2}})}^{2} = {(k - 1)}^{2}$

Simplify the left side.

Simplify ${({(3 k - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 k - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(3 k - 5)}^{\frac{1}{2} \cdot 2} = {(k - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 k - 5)}^{\frac{1}{2} \cdot 2} = {(k - 1)}^{2}$

Rewrite the expression.

${(3 k - 5)}^{1} = {(k - 1)}^{2}$

Simplify.

$3 k - 5 = {(k - 1)}^{2}$

Simplify the right side.

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

Rewrite ${(k - 1)}^{2}$ as $(k - 1) (k - 1)$ .

$3 k - 5 = (k - 1) (k - 1)$

Expand $(k - 1) (k - 1)$ using the FOIL Method.

Apply the distributive property.

$3 k - 5 = k (k - 1) - 1 (k - 1)$

Apply the distributive property.

$3 k - 5 = k \cdot k + k \cdot - 1 - 1 (k - 1)$

Apply the distributive property.

$3 k - 5 = k \cdot k + k \cdot - 1 - 1 k - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $k$ by $k$ .

$3 k - 5 = k^{2} + k \cdot - 1 - 1 k - 1 \cdot - 1$

Move $- 1$ to the left of $k$ .

$3 k - 5 = k^{2} - 1 \cdot k - 1 k - 1 \cdot - 1$

Rewrite $- 1 k$ as $- k$ .

$3 k - 5 = k^{2} - k - 1 k - 1 \cdot - 1$

Rewrite $- 1 k$ as $- k$ .

$3 k - 5 = k^{2} - k - k - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$3 k - 5 = k^{2} - k - k + 1$

Subtract $k$ from $- k$ .

$3 k - 5 = k^{2} - 2 k + 1$

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.

$k^{2} - 2 k + 1 = 3 k - 5$

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

Subtract $3 k$ from both sides of the equation.

$k^{2} - 2 k + 1 - 3 k = - 5$

Subtract $3 k$ from $- 2 k$ .

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

Add $5$ to both sides of the equation.

$k^{2} - 5 k + 1 + 5 = 0$

Add $1$ and $5$ .

$k^{2} - 5 k + 6 = 0$

Factor $k^{2} - 5 k + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(k - 3) (k - 2) = 0$

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

$k - 3 = 0$

$k - 2 = 0$

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

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

$k - 3 = 0$

Add $3$ to both sides of the equation.

$k = 3$

Set $k - 2$ equal to $0$ and solve for $k$ .

Set $k - 2$ equal to $0$ .

$k - 2 = 0$

Add $2$ to both sides of the equation.

$k = 2$

The final solution is all the values that make $(k - 3) (k - 2) = 0$ true.

$k = 3, 2$

Solve for k k=1+ square root of 3k-5

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