Solve by Factoring x+1 = square root of 19-x

$x + 1 = \sqrt{19 - x}$

Step 1

Subtract $\sqrt{19 - x}$ from both sides of the equation.

$x + 1 - \sqrt{19 - x} = 0$

Step 2

Move all terms not containing $- \sqrt{19 - x}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$1 - \sqrt{19 - x} = - x$

Subtract $1$ from both sides of the equation.

$- \sqrt{19 - x} = - x - 1$

Step 3

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

${(- \sqrt{19 - x})}^{2} = {(- x - 1)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{19 - x}$ as ${(19 - x)}^{\frac{1}{2}}$ .

${(- {(19 - x)}^{\frac{1}{2}})}^{2} = {(- x - 1)}^{2}$

Simplify the left side.

Simplify ${(- {(19 - x)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(19 - x)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(19 - x)}^{\frac{1}{2}})}^{2} = {(- x - 1)}^{2}$

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

$1 {({(19 - x)}^{\frac{1}{2}})}^{2} = {(- x - 1)}^{2}$

Multiply ${({(19 - x)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(19 - x)}^{\frac{1}{2}})}^{2} = {(- x - 1)}^{2}$

Multiply the exponents in ${({(19 - x)}^{\frac{1}{2}})}^{2}$ .

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

${(19 - x)}^{\frac{1}{2} \cdot 2} = {(- x - 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(19 - x)}^{\frac{1}{2} \cdot 2} = {(- x - 1)}^{2}$

Rewrite the expression.

${(19 - x)}^{1} = {(- x - 1)}^{2}$

Simplify.

$19 - x = {(- x - 1)}^{2}$

Simplify the right side.

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

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

$19 - x = (- x - 1) (- x - 1)$

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

Apply the distributive property.

$19 - x = - x (- x - 1) - 1 (- x - 1)$

Apply the distributive property.

$19 - x = - x (- x) - x \cdot - 1 - 1 (- x - 1)$

Apply the distributive property.

$19 - x = - x (- x) - x \cdot - 1 - 1 (- x) - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$19 - x = - 1 \cdot - 1 x \cdot x - x \cdot - 1 - 1 (- x) - 1 \cdot - 1$

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

Move $x$ .

$19 - x = - 1 \cdot - 1 (x \cdot x) - x \cdot - 1 - 1 (- x) - 1 \cdot - 1$

Multiply $x$ by $x$ .

$19 - x = - 1 \cdot - 1 x^{2} - x \cdot - 1 - 1 (- x) - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$19 - x = 1 x^{2} - x \cdot - 1 - 1 (- x) - 1 \cdot - 1$

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

$19 - x = x^{2} - x \cdot - 1 - 1 (- x) - 1 \cdot - 1$

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$19 - x = x^{2} + 1 x - 1 (- x) - 1 \cdot - 1$

Multiply $x$ by $1$ .

$19 - x = x^{2} + x - 1 (- x) - 1 \cdot - 1$

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

$19 - x = x^{2} + x + 1 x - 1 \cdot - 1$

Multiply $x$ by $1$ .

$19 - x = x^{2} + x + x - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$19 - x = x^{2} + x + x + 1$

Add $x$ and $x$ .

$19 - x = x^{2} + 2 x + 1$

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.

$x^{2} + 2 x + 1 = 19 - x$

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

Add $x$ to both sides of the equation.

$x^{2} + 2 x + 1 + x = 19$

Add $2 x$ and $x$ .

$x^{2} + 3 x + 1 = 19$

Subtract $19$ from both sides of the equation.

$x^{2} + 3 x + 1 - 19 = 0$

Subtract $19$ from $1$ .

$x^{2} + 3 x - 18 = 0$

Factor $x^{2} + 3 x - 18$ 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 $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$(x - 3) (x + 6) = 0$

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

$x - 3 = 0$

$x + 6 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x - 3) (x + 6) = 0$ true.

$x = 3, - 6$

Step 6

Exclude the solutions that do not make $x + 1 - \sqrt{19 - x} = 0$ true.

$x = 3$

Solve by Factoring x+1 = square root of 19-x

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