Find the Roots (Zeros) square root of x-1=x-3

$\sqrt{x - 1} = x - 3$

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$x - 1 = (x - 3) (x - 3)$

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

Apply the distributive property.

$x - 1 = x (x - 3) - 3 (x - 3)$

Apply the distributive property.

$x - 1 = x \cdot x + x \cdot - 3 - 3 (x - 3)$

Apply the distributive property.

$x - 1 = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

$x - 1 = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$x - 1 = x^{2} - 3 x - 3 x + 9$

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

$x - 1 = x^{2} - 6 x + 9$

Step 3

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.

$x^{2} - 6 x + 9 = x - 1$

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

Subtract $x$ from both sides of the equation.

$x^{2} - 6 x + 9 - x = - 1$

Subtract $x$ from $- 6 x$ .

$x^{2} - 7 x + 9 = - 1$

Add $1$ to both sides of the equation.

$x^{2} - 7 x + 9 + 1 = 0$

Add $9$ and $1$ .

$x^{2} - 7 x + 10 = 0$

Factor $x^{2} - 7 x + 10$ 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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

$(x - 5) (x - 2) = 0$

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

$x - 5 = 0$

$x - 2 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x - 5) (x - 2) = 0$ true.

$x = 5, 2$

Step 4

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

$x = 5$

Step 5

Find the Roots (Zeros) square root of x-1=x-3

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