Solve by Factoring square root of 3x+13=1+ square root of x+4

$\sqrt{3 x + 13} = 1 + \sqrt{x + 4}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $1$ from both sides of the equation.

$\sqrt{3 x + 13} - 1 = \sqrt{x + 4}$

Subtract $\sqrt{x + 4}$ from both sides of the equation.

$\sqrt{3 x + 13} - 1 - \sqrt{x + 4} = 0$

Step 2

Move all terms not containing $\sqrt{3 x + 13}$ to the right side of the equation.

Add $1$ to both sides of the equation.

$\sqrt{3 x + 13} - \sqrt{x + 4} = 1$

Add $\sqrt{x + 4}$ to both sides of the equation.

$\sqrt{3 x + 13} = 1 + \sqrt{x + 4}$

Step 3

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

${\sqrt{3 x + 13}}^{2} = {(1 + \sqrt{x + 4})}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 13}$ as ${(3 x + 13)}^{\frac{1}{2}}$ .

${({(3 x + 13)}^{\frac{1}{2}})}^{2} = {(1 + \sqrt{x + 4})}^{2}$

Simplify the left side.

Simplify ${({(3 x + 13)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 x + 13)}^{\frac{1}{2}})}^{2}$ .

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

${(3 x + 13)}^{\frac{1}{2} \cdot 2} = {(1 + \sqrt{x + 4})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x + 13)}^{\frac{1}{2} \cdot 2} = {(1 + \sqrt{x + 4})}^{2}$

Rewrite the expression.

${(3 x + 13)}^{1} = {(1 + \sqrt{x + 4})}^{2}$

Simplify.

$3 x + 13 = {(1 + \sqrt{x + 4})}^{2}$

Simplify the right side.

Simplify ${(1 + \sqrt{x + 4})}^{2}$ .

Rewrite ${(1 + \sqrt{x + 4})}^{2}$ as $(1 + \sqrt{x + 4}) (1 + \sqrt{x + 4})$ .

$3 x + 13 = (1 + \sqrt{x + 4}) (1 + \sqrt{x + 4})$

Expand $(1 + \sqrt{x + 4}) (1 + \sqrt{x + 4})$ using the FOIL Method.

Apply the distributive property.

$3 x + 13 = 1 (1 + \sqrt{x + 4}) + \sqrt{x + 4} (1 + \sqrt{x + 4})$

Apply the distributive property.

$3 x + 13 = 1 \cdot 1 + 1 \sqrt{x + 4} + \sqrt{x + 4} (1 + \sqrt{x + 4})$

Apply the distributive property.

$3 x + 13 = 1 \cdot 1 + 1 \sqrt{x + 4} + \sqrt{x + 4} \cdot 1 + \sqrt{x + 4} \sqrt{x + 4}$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$3 x + 13 = 1 + 1 \sqrt{x + 4} + \sqrt{x + 4} \cdot 1 + \sqrt{x + 4} \sqrt{x + 4}$

Multiply $\sqrt{x + 4}$ by $1$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} \cdot 1 + \sqrt{x + 4} \sqrt{x + 4}$

Multiply $\sqrt{x + 4}$ by $1$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + \sqrt{x + 4} \sqrt{x + 4}$

Multiply $\sqrt{x + 4} \sqrt{x + 4}$ .

Raise $\sqrt{x + 4}$ to the power of $1$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {\sqrt{x + 4}}^{1} \sqrt{x + 4}$

Raise $\sqrt{x + 4}$ to the power of $1$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {\sqrt{x + 4}}^{1} {\sqrt{x + 4}}^{1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {\sqrt{x + 4}}^{1 + 1}$

Add $1$ and $1$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {\sqrt{x + 4}}^{2}$

Rewrite ${\sqrt{x + 4}}^{2}$ as $x + 4$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 4}$ as ${(x + 4)}^{\frac{1}{2}}$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {({(x + 4)}^{\frac{1}{2}})}^{2}$

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

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {(x + 4)}^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {(x + 4)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {(x + 4)}^{\frac{2}{2}}$

Rewrite the expression.

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + {(x + 4)}^{1}$

Simplify.

$3 x + 13 = 1 + \sqrt{x + 4} + \sqrt{x + 4} + x + 4$

Add $1$ and $4$ .

$3 x + 13 = 5 + \sqrt{x + 4} + \sqrt{x + 4} + x$

Add $\sqrt{x + 4}$ and $\sqrt{x + 4}$ .

$3 x + 13 = 5 + 2 \sqrt{x + 4} + x$

Step 5

Solve for $2 \sqrt{x + 4}$ .

Rewrite the equation as $5 + 2 \sqrt{x + 4} + x = 3 x + 13$ .

$5 + 2 \sqrt{x + 4} + x = 3 x + 13$

Move all terms not containing $2 \sqrt{x + 4}$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$2 \sqrt{x + 4} + x = 3 x + 13 - 5$

Subtract $x$ from both sides of the equation.

$2 \sqrt{x + 4} = 3 x + 13 - 5 - x$

Subtract $x$ from $3 x$ .

$2 \sqrt{x + 4} = 2 x + 13 - 5$

Subtract $5$ from $13$ .

$2 \sqrt{x + 4} = 2 x + 8$

Step 6

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

${(2 \sqrt{x + 4})}^{2} = {(2 x + 8)}^{2}$

Step 7

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 4}$ as ${(x + 4)}^{\frac{1}{2}}$ .

${(2 {(x + 4)}^{\frac{1}{2}})}^{2} = {(2 x + 8)}^{2}$

Simplify the left side.

Simplify ${(2 {(x + 4)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(x + 4)}^{\frac{1}{2}}$ .

$2^{2} {({(x + 4)}^{\frac{1}{2}})}^{2} = {(2 x + 8)}^{2}$

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

$4 {({(x + 4)}^{\frac{1}{2}})}^{2} = {(2 x + 8)}^{2}$

Multiply the exponents in ${({(x + 4)}^{\frac{1}{2}})}^{2}$ .

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

$4 {(x + 4)}^{\frac{1}{2} \cdot 2} = {(2 x + 8)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(x + 4)}^{\frac{1}{2} \cdot 2} = {(2 x + 8)}^{2}$

Rewrite the expression.

$4 {(x + 4)}^{1} = {(2 x + 8)}^{2}$

Simplify.

$4 (x + 4) = {(2 x + 8)}^{2}$

Apply the distributive property.

$4 x + 4 \cdot 4 = {(2 x + 8)}^{2}$

Multiply $4$ by $4$ .

$4 x + 16 = {(2 x + 8)}^{2}$

Simplify the right side.

Simplify ${(2 x + 8)}^{2}$ .

Rewrite ${(2 x + 8)}^{2}$ as $(2 x + 8) (2 x + 8)$ .

$4 x + 16 = (2 x + 8) (2 x + 8)$

Expand $(2 x + 8) (2 x + 8)$ using the FOIL Method.

Apply the distributive property.

$4 x + 16 = 2 x (2 x + 8) + 8 (2 x + 8)$

Apply the distributive property.

$4 x + 16 = 2 x (2 x) + 2 x \cdot 8 + 8 (2 x + 8)$

Apply the distributive property.

$4 x + 16 = 2 x (2 x) + 2 x \cdot 8 + 8 (2 x) + 8 \cdot 8$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$4 x + 16 = 2 \cdot 2 x \cdot x + 2 x \cdot 8 + 8 (2 x) + 8 \cdot 8$

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

Move $x$ .

$4 x + 16 = 2 \cdot 2 (x \cdot x) + 2 x \cdot 8 + 8 (2 x) + 8 \cdot 8$

Multiply $x$ by $x$ .

$4 x + 16 = 2 \cdot 2 x^{2} + 2 x \cdot 8 + 8 (2 x) + 8 \cdot 8$

Multiply $2$ by $2$ .

$4 x + 16 = 4 x^{2} + 2 x \cdot 8 + 8 (2 x) + 8 \cdot 8$

Multiply $8$ by $2$ .

$4 x + 16 = 4 x^{2} + 16 x + 8 (2 x) + 8 \cdot 8$

Multiply $2$ by $8$ .

$4 x + 16 = 4 x^{2} + 16 x + 16 x + 8 \cdot 8$

Multiply $8$ by $8$ .

$4 x + 16 = 4 x^{2} + 16 x + 16 x + 64$

Add $16 x$ and $16 x$ .

$4 x + 16 = 4 x^{2} + 32 x + 64$

Step 8

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.

$4 x^{2} + 32 x + 64 = 4 x + 16$

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

Subtract $4 x$ from both sides of the equation.

$4 x^{2} + 32 x + 64 - 4 x = 16$

Subtract $4 x$ from $32 x$ .

$4 x^{2} + 28 x + 64 = 16$

Subtract $16$ from both sides of the equation.

$4 x^{2} + 28 x + 64 - 16 = 0$

Subtract $16$ from $64$ .

$4 x^{2} + 28 x + 48 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 x^{2} + 28 x + 48$ .

Factor $4$ out of $4 x^{2}$ .

$4 (x^{2}) + 28 x + 48 = 0$

Factor $4$ out of $28 x$ .

$4 (x^{2}) + 4 (7 x) + 48 = 0$

Factor $4$ out of $48$ .

$4 x^{2} + 4 (7 x) + 4 \cdot 12 = 0$

Factor $4$ out of $4 x^{2} + 4 (7 x)$ .

$4 (x^{2} + 7 x) + 4 \cdot 12 = 0$

Factor $4$ out of $4 (x^{2} + 7 x) + 4 \cdot 12$ .

$4 (x^{2} + 7 x + 12) = 0$

Factor.

Factor $x^{2} + 7 x + 12$ 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 $12$ and whose sum is $7$ .

$3, 4$

Write the factored form using these integers.

$4 ((x + 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$4 (x + 3) (x + 4) = 0$

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

$x + 3 = 0$

$x + 4 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $4 (x + 3) (x + 4) = 0$ true.

$x = - 3, - 4$

Solve by Factoring square root of 3x+13=1+ square root of x+4

Related Questions

Support LCMGCF.com by disabling your adblocker.