Solve for x square root of 2x+6+4=x+3

$\sqrt{2 x + 6} + 4 = x + 3$

Move all terms not containing $\sqrt{2 x + 6}$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$\sqrt{2 x + 6} = x + 3 - 4$

Subtract $4$ from $3$ .

$\sqrt{2 x + 6} = x - 1$

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

${\sqrt{2 x + 6}}^{2} = {(x - 1)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 6}$ as ${(2 x + 6)}^{\frac{1}{2}}$ .

${({(2 x + 6)}^{\frac{1}{2}})}^{2} = {(x - 1)}^{2}$

Simplify the left side.

Simplify ${({(2 x + 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 x + 6)}^{\frac{1}{2}})}^{2}$ .

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

${(2 x + 6)}^{\frac{1}{2} \cdot 2} = {(x - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x + 6)}^{\frac{1}{2} \cdot 2} = {(x - 1)}^{2}$

Rewrite the expression.

${(2 x + 6)}^{1} = {(x - 1)}^{2}$

Simplify.

$2 x + 6 = {(x - 1)}^{2}$

Simplify the right side.

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

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

$2 x + 6 = (x - 1) (x - 1)$

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

Apply the distributive property.

$2 x + 6 = x (x - 1) - 1 (x - 1)$

Apply the distributive property.

$2 x + 6 = x \cdot x + x \cdot - 1 - 1 (x - 1)$

Apply the distributive property.

$2 x + 6 = x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$2 x + 6 = x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1$

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

$2 x + 6 = x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1$

Rewrite $- 1 x$ as $- x$ .

$2 x + 6 = x^{2} - x - 1 x - 1 \cdot - 1$

Rewrite $- 1 x$ as $- x$ .

$2 x + 6 = x^{2} - x - x - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $- x$ .

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

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.

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

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

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

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

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

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

Subtract $6$ from both sides of the equation.

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

Subtract $6$ from $1$ .

$x^{2} - 4 x - 5 = 0$

Factor $x^{2} - 4 x - 5$ 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 $- 4$ .

$- 5, 1$

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 $0$ .

$x + 1 = 0$

Subtract $1$ from 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 $\sqrt{2 x + 6} + 4 = x + 3$ true.

$x = 5$

Solve for x square root of 2x+6+4=x+3

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