Solve sqrt{12-n}=n

$$\sqrt { 12 - n } = n$$

Solve for n

$$n=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{12-n}\right)^{2}=n^{2}$$

Calculate \sqrt{12-n} to the power of 2 and get 12-n.

$$12-n=n^{2}$$

Subtract n^{2} from both sides.

$$12-n-n^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-n^{2}-n+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn+12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=3 b=-4$$

Rewrite -n^{2}-n+12 as \left(-n^{2}+3n\right)+\left(-4n+12\right).

$$\left(-n^{2}+3n\right)+\left(-4n+12\right)$$

Factor out n in the first and 4 in the second group.

$$n\left(-n+3\right)+4\left(-n+3\right)$$

Factor out common term -n+3 by using distributive property.

$$\left(-n+3\right)\left(n+4\right)$$

To find equation solutions, solve -n+3=0 and n+4=0.

$$n=3 n=-4$$

Substitute 3 for n in the equation \sqrt{12-n}=n.

$$\sqrt{12-3}=3$$

Simplify. The value n=3 satisfies the equation.

$$3=3$$

Substitute -4 for n in the equation \sqrt{12-n}=n.

$$\sqrt{12-\left(-4\right)}=-4$$

Simplify. The value n=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-4$$

Equation \sqrt{12-n}=n has a unique solution.

$$n=3$$