Find the Roots (Zeros) square root of x+1=x-1

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

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

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

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

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $- x$ .

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

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.

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

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

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $- 2 x$ .

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

Subtract $1$ from both sides of the equation.

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

Combine the opposite terms in $x^{2} - 3 x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $x^{2} - 3 x$ and $0$ .

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

Factor $x$ out of $x^{2} - 3 x$ .

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

$x \cdot x - 3 x = 0$

Factor $x$ out of $- 3 x$ .

$x \cdot x + x \cdot - 3 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 3$ .

$x (x - 3) = 0$

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

$x = 0$

$x - 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 3$

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

$x = 0, 3$

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

$x = 3$

Find the Roots (Zeros) square root of x+1=x-1

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