Solve for x x+3 = square root of 2x-1+5

$x + 3 = \sqrt{2 x - 1} + 5$

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

$\sqrt{2 x - 1} + 5 = x + 3$

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

Subtract $5$ from both sides of the equation.

$\sqrt{2 x - 1} = x + 3 - 5$

Subtract $5$ from $3$ .

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

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

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

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

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.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 2$ to the left of $x$ .

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

Multiply $- 2$ by $- 2$ .

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

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

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

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.

$x = 5, 1$

Exclude the solutions that do not make $x + 3 = \sqrt{2 x - 1} + 5$ true.

$x = 5$

Solve for x x+3 = square root of 2x-1+5

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