Solve for x square root of x+5+1=x

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

Step 1

Subtract $1$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$x + 5 = (x - 1) (x - 1)$

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

Apply the distributive property.

$x + 5 = x (x - 1) - 1 (x - 1)$

Apply the distributive property.

$x + 5 = x \cdot x + x \cdot - 1 - 1 (x - 1)$

Apply the distributive property.

$x + 5 = x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

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

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

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

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

Multiply $- 1$ by $- 1$ .

$x + 5 = x^{2} - x - x + 1$

Subtract $x$ from $- x$ .

$x + 5 = x^{2} - 2 x + 1$

Step 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.

$x^{2} - 2 x + 1 = x + 5$

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

Subtract $x$ from both sides of the equation.

$x^{2} - 2 x + 1 - x = 5$

Subtract $x$ from $- 2 x$ .

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

Subtract $5$ from both sides of the equation.

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

Subtract $5$ from $1$ .

$x^{2} - 3 x - 4 = 0$

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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 $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 5

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

$x = 4$

Solve for x square root of x+5+1=x

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