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

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

Step 1

Subtract $1$ from both sides of the equation.

$\sqrt{2 x + 3} - \sqrt{x + 5} - 1 = 0$

Step 2

Move all terms not containing $\sqrt{2 x + 3}$ to the right side of the equation.

Add $\sqrt{x + 5}$ to both sides of the equation.

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

Add $1$ to both sides of the equation.

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

Step 3

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

${\sqrt{2 x + 3}}^{2} = {(\sqrt{x + 5} + 1)}^{2}$

Step 4

Simplify each side of the equation.

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

${({(2 x + 3)}^{\frac{1}{2}})}^{2} = {(\sqrt{x + 5} + 1)}^{2}$

Simplify the left side.

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

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

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

${(2 x + 3)}^{\frac{1}{2} \cdot 2} = {(\sqrt{x + 5} + 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x + 3)}^{\frac{1}{2} \cdot 2} = {(\sqrt{x + 5} + 1)}^{2}$

Rewrite the expression.

${(2 x + 3)}^{1} = {(\sqrt{x + 5} + 1)}^{2}$

Simplify.

$2 x + 3 = {(\sqrt{x + 5} + 1)}^{2}$

Simplify the right side.

Simplify ${(\sqrt{x + 5} + 1)}^{2}$ .

Rewrite ${(\sqrt{x + 5} + 1)}^{2}$ as $(\sqrt{x + 5} + 1) (\sqrt{x + 5} + 1)$ .

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

Expand $(\sqrt{x + 5} + 1) (\sqrt{x + 5} + 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$2 x + 3 = \sqrt{x + 5} \sqrt{x + 5} + \sqrt{x + 5} \cdot 1 + 1 (\sqrt{x + 5} + 1)$

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{x + 5} \sqrt{x + 5}$ .

Raise $\sqrt{x + 5}$ to the power of $1$ .

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

Raise $\sqrt{x + 5}$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{x + 5}}^{2}$ as $x + 5$ .

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

$2 x + 3 = {({(x + 5)}^{\frac{1}{2}})}^{2} + \sqrt{x + 5} \cdot 1 + 1 \sqrt{x + 5} + 1 \cdot 1$

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

$2 x + 3 = {(x + 5)}^{\frac{1}{2} \cdot 2} + \sqrt{x + 5} \cdot 1 + 1 \sqrt{x + 5} + 1 \cdot 1$

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

$2 x + 3 = {(x + 5)}^{\frac{2}{2}} + \sqrt{x + 5} \cdot 1 + 1 \sqrt{x + 5} + 1 \cdot 1$

Cancel the common factor of $2$ .

Cancel the common factor.

$2 x + 3 = {(x + 5)}^{\frac{2}{2}} + \sqrt{x + 5} \cdot 1 + 1 \sqrt{x + 5} + 1 \cdot 1$

Rewrite the expression.

$2 x + 3 = {(x + 5)}^{1} + \sqrt{x + 5} \cdot 1 + 1 \sqrt{x + 5} + 1 \cdot 1$

Simplify.

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

Multiply $\sqrt{x + 5}$ by $1$ .

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

Multiply $\sqrt{x + 5}$ by $1$ .

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

Multiply $1$ by $1$ .

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

Add $5$ and $1$ .

$2 x + 3 = x + 6 + \sqrt{x + 5} + \sqrt{x + 5}$

Add $\sqrt{x + 5}$ and $\sqrt{x + 5}$ .

$2 x + 3 = x + 6 + 2 \sqrt{x + 5}$

Step 5

Solve for $2 \sqrt{x + 5}$ .

Rewrite the equation as $x + 6 + 2 \sqrt{x + 5} = 2 x + 3$ .

$x + 6 + 2 \sqrt{x + 5} = 2 x + 3$

Move all terms not containing $2 \sqrt{x + 5}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$6 + 2 \sqrt{x + 5} = 2 x + 3 - x$

Subtract $6$ from both sides of the equation.

$2 \sqrt{x + 5} = 2 x + 3 - x - 6$

Subtract $x$ from $2 x$ .

$2 \sqrt{x + 5} = x + 3 - 6$

Subtract $6$ from $3$ .

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

Step 6

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

${(2 \sqrt{x + 5})}^{2} = {(x - 3)}^{2}$

Step 7

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${(2 {(x + 5)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(x + 5)}^{\frac{1}{2}}$ .

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

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

$4 {({(x + 5)}^{\frac{1}{2}})}^{2} = {(x - 3)}^{2}$

Multiply the exponents in ${({(x + 5)}^{\frac{1}{2}})}^{2}$ .

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

$4 {(x + 5)}^{\frac{1}{2} \cdot 2} = {(x - 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(x + 5)}^{\frac{1}{2} \cdot 2} = {(x - 3)}^{2}$

Rewrite the expression.

$4 {(x + 5)}^{1} = {(x - 3)}^{2}$

Simplify.

$4 (x + 5) = {(x - 3)}^{2}$

Apply the distributive property.

$4 x + 4 \cdot 5 = {(x - 3)}^{2}$

Multiply $4$ by $5$ .

$4 x + 20 = {(x - 3)}^{2}$

Simplify the right side.

Simplify ${(x - 3)}^{2}$ .

Rewrite ${(x - 3)}^{2}$ as $(x - 3) (x - 3)$ .

$4 x + 20 = (x - 3) (x - 3)$

Expand $(x - 3) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$4 x + 20 = x (x - 3) - 3 (x - 3)$

Apply the distributive property.

$4 x + 20 = x \cdot x + x \cdot - 3 - 3 (x - 3)$

Apply the distributive property.

$4 x + 20 = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$4 x + 20 = x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3$

Move $- 3$ to the left of $x$ .

$4 x + 20 = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$4 x + 20 = x^{2} - 3 x - 3 x + 9$

Subtract $3 x$ from $- 3 x$ .

$4 x + 20 = x^{2} - 6 x + 9$

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.

$x^{2} - 6 x + 9 = 4 x + 20$

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

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

$x^{2} - 6 x + 9 - 4 x = 20$

Subtract $4 x$ from $- 6 x$ .

$x^{2} - 10 x + 9 = 20$

Subtract $20$ from both sides of the equation.

$x^{2} - 10 x + 9 - 20 = 0$

Subtract $20$ from $9$ .

$x^{2} - 10 x - 11 = 0$

Factor $x^{2} - 10 x - 11$ 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 $- 11$ and whose sum is $- 10$ .

$- 11, 1$

Write the factored form using these integers.

$(x - 11) (x + 1) = 0$

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

$x - 11 = 0$

$x + 1 = 0$

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

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 $(x - 11) (x + 1) = 0$ true.

$x = 11, - 1$

Step 9

Exclude the solutions that do not make $\sqrt{2 x + 3} - \sqrt{x + 5} - 1 = 0$ true.

$x = 11$

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

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