Solve sqrt{72-k}=k

$$\sqrt { 72 - k } = k$$

Solve for k

$$k=8$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{72-k}\right)^{2}=k^{2}$$

Calculate \sqrt{72-k} to the power of 2 and get 72-k.

$$72-k=k^{2}$$

Subtract k^{2} from both sides.

$$72-k-k^{2}=0$$

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

$$-k^{2}-k+72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -k^{2}+ak+bk+72. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-72=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=8 b=-9$$

Rewrite -k^{2}-k+72 as \left(-k^{2}+8k\right)+\left(-9k+72\right).

$$\left(-k^{2}+8k\right)+\left(-9k+72\right)$$

Factor out k in the first and 9 in the second group.

$$k\left(-k+8\right)+9\left(-k+8\right)$$

Factor out common term -k+8 by using distributive property.

$$\left(-k+8\right)\left(k+9\right)$$

To find equation solutions, solve -k+8=0 and k+9=0.

$$k=8 k=-9$$

Substitute 8 for k in the equation \sqrt{72-k}=k.

$$\sqrt{72-8}=8$$

Simplify. The value k=8 satisfies the equation.

$$8=8$$

Substitute -9 for k in the equation \sqrt{72-k}=k.

$$\sqrt{72-\left(-9\right)}=-9$$

Simplify. The value k=-9 does not satisfy the equation because the left and the right hand side have opposite signs.

$$9=-9$$

Equation \sqrt{72-k}=k has a unique solution.

$$k=8$$