Solve for n square root of n-1=n-7

$\sqrt{n - 1} = n - 7$

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

${\sqrt{n - 1}}^{2} = {(n - 7)}^{2}$

Simplify each side of the equation.

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

${({(n - 1)}^{\frac{1}{2}})}^{2} = {(n - 7)}^{2}$

Simplify the left side.

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

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

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

${(n - 1)}^{\frac{1}{2} \cdot 2} = {(n - 7)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(n - 1)}^{\frac{1}{2} \cdot 2} = {(n - 7)}^{2}$

Rewrite the expression.

${(n - 1)}^{1} = {(n - 7)}^{2}$

Simplify.

$n - 1 = {(n - 7)}^{2}$

Simplify the right side.

Simplify ${(n - 7)}^{2}$ .

Rewrite ${(n - 7)}^{2}$ as $(n - 7) (n - 7)$ .

$n - 1 = (n - 7) (n - 7)$

Expand $(n - 7) (n - 7)$ using the FOIL Method.

Apply the distributive property.

$n - 1 = n (n - 7) - 7 (n - 7)$

Apply the distributive property.

$n - 1 = n \cdot n + n \cdot - 7 - 7 (n - 7)$

Apply the distributive property.

$n - 1 = n \cdot n + n \cdot - 7 - 7 n - 7 \cdot - 7$

Simplify and combine like terms.

Simplify each term.

Multiply $n$ by $n$ .

$n - 1 = n^{2} + n \cdot - 7 - 7 n - 7 \cdot - 7$

Move $- 7$ to the left of $n$ .

$n - 1 = n^{2} - 7 \cdot n - 7 n - 7 \cdot - 7$

Multiply $- 7$ by $- 7$ .

$n - 1 = n^{2} - 7 n - 7 n + 49$

Subtract $7 n$ from $- 7 n$ .

$n - 1 = n^{2} - 14 n + 49$

Solve for $n$ .

Since $n$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$n^{2} - 14 n + 49 = n - 1$

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

Subtract $n$ from both sides of the equation.

$n^{2} - 14 n + 49 - n = - 1$

Subtract $n$ from $- 14 n$ .

$n^{2} - 15 n + 49 = - 1$

Add $1$ to both sides of the equation.

$n^{2} - 15 n + 49 + 1 = 0$

Add $49$ and $1$ .

$n^{2} - 15 n + 50 = 0$

Factor $n^{2} - 15 n + 50$ 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 $50$ and whose sum is $- 15$ .

$- 10, - 5$

Write the factored form using these integers.

$(n - 10) (n - 5) = 0$

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

$n - 10 = 0$

$n - 5 = 0$

Set $n - 10$ equal to $0$ and solve for $n$ .

Set $n - 10$ equal to $0$ .

$n - 10 = 0$

Add $10$ to both sides of the equation.

$n = 10$

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

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

$n - 5 = 0$

Add $5$ to both sides of the equation.

$n = 5$

The final solution is all the values that make $(n - 10) (n - 5) = 0$ true.

$n = 10, 5$

Exclude the solutions that do not make $\sqrt{n - 1} = n - 7$ true.

$n = 10$

Solve for n square root of n-1=n-7

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