Solve for n 3+2 square root of n=n

$3 + 2 \sqrt{n} = n$

Subtract $3$ from both sides of the equation.

$2 \sqrt{n} = n - 3$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $2 n^{\frac{1}{2}}$ .

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

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

$4 {(n^{\frac{1}{2}})}^{2} = {(n - 3)}^{2}$

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

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

$4 n^{\frac{1}{2} \cdot 2} = {(n - 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 n^{\frac{1}{2} \cdot 2} = {(n - 3)}^{2}$

Rewrite the expression.

$4 n^{1} = {(n - 3)}^{2}$

Simplify.

$4 n = {(n - 3)}^{2}$

Simplify the right side.

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

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

$4 n = (n - 3) (n - 3)$

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

Apply the distributive property.

$4 n = n (n - 3) - 3 (n - 3)$

Apply the distributive property.

$4 n = n \cdot n + n \cdot - 3 - 3 (n - 3)$

Apply the distributive property.

$4 n = n \cdot n + n \cdot - 3 - 3 n - 3 \cdot - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $n$ by $n$ .

$4 n = n^{2} + n \cdot - 3 - 3 n - 3 \cdot - 3$

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

$4 n = n^{2} - 3 \cdot n - 3 n - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$4 n = n^{2} - 3 n - 3 n + 9$

Subtract $3 n$ from $- 3 n$ .

$4 n = n^{2} - 6 n + 9$

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.

$n^{2} - 6 n + 9 = 4 n$

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

Subtract $4 n$ from both sides of the equation.

$n^{2} - 6 n + 9 - 4 n = 0$

Subtract $4 n$ from $- 6 n$ .

$n^{2} - 10 n + 9 = 0$

Factor $n^{2} - 10 n + 9$ 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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Write the factored form using these integers.

$(n - 9) (n - 1) = 0$

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

$n - 9 = 0$

$n - 1 = 0$

Set $n - 9$ equal to $0$ and solve for $n$ .

Set $n - 9$ equal to $0$ .

$n - 9 = 0$

Add $9$ to both sides of the equation.

$n = 9$

Set $n - 1$ equal to $0$ and solve for $n$ .

Set $n - 1$ equal to $0$ .

$n - 1 = 0$

Add $1$ to both sides of the equation.

$n = 1$

The final solution is all the values that make $(n - 9) (n - 1) = 0$ true.

$n = 9, 1$

Exclude the solutions that do not make $3 + 2 \sqrt{n} = n$ true.

$n = 9$

Solve for n 3+2 square root of n=n

Related Questions

Support LCMGCF.com by disabling your adblocker.