Solve by Factoring x-2 = square root of 2x-1

$x - 2 = \sqrt{2 x - 1}$

Step 1

Subtract $\sqrt{2 x - 1}$ from both sides of the equation.

$x - 2 - \sqrt{2 x - 1} = 0$

Step 2

Move all terms not containing $- \sqrt{2 x - 1}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$- 2 - \sqrt{2 x - 1} = - x$

Add $2$ to both sides of the equation.

$- \sqrt{2 x - 1} = - x + 2$

Step 3

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

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

Step 4

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- {(2 x - 1)}^{\frac{1}{2}}$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply ${({(2 x - 1)}^{\frac{1}{2}})}^{2}$ by $1$ .

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$2 x - 1 = (- x + 2) (- x + 2)$

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

Apply the distributive property.

$2 x - 1 = - x (- x + 2) + 2 (- x + 2)$

Apply the distributive property.

$2 x - 1 = - x (- x) - x \cdot 2 + 2 (- x + 2)$

Apply the distributive property.

$2 x - 1 = - x (- x) - x \cdot 2 + 2 (- x) + 2 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 x - 1 = - 1 \cdot - 1 x \cdot x - x \cdot 2 + 2 (- x) + 2 \cdot 2$

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

Move $x$ .

$2 x - 1 = - 1 \cdot - 1 (x \cdot x) - x \cdot 2 + 2 (- x) + 2 \cdot 2$

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

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

Multiply $2$ by $2$ .

$2 x - 1 = x^{2} - 2 x - 2 x + 4$

Subtract $2 x$ from $- 2 x$ .

$2 x - 1 = x^{2} - 4 x + 4$

Step 5

Solve for $x$ .

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

$x^{2} - 4 x + 4 = 2 x - 1$

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

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

$x^{2} - 4 x + 4 - 2 x = - 1$

Subtract $2 x$ from $- 4 x$ .

$x^{2} - 6 x + 4 = - 1$

Add $1$ to both sides of the equation.

$x^{2} - 6 x + 4 + 1 = 0$

Add $4$ and $1$ .

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

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

$- 5, - 1$

Write the factored form using these integers.

$(x - 5) (x - 1) = 0$

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

$x - 5 = 0$

$x - 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 5) (x - 1) = 0$ true.

$x = 5, 1$

Step 6

Exclude the solutions that do not make $x - 2 - \sqrt{2 x - 1} = 0$ true.

$x = 5$

Solve by Factoring x-2 = square root of 2x-1

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