Solve sqrt{90-n}=n

$$\sqrt { 90 - n } = n$$

Solve for n

$$n=9$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{90-n}\right)^{2}=n^{2}$$

Calculate \sqrt{90-n} to the power of 2 and get 90-n.

$$90-n=n^{2}$$

Subtract n^{2} from both sides.

$$90-n-n^{2}=0$$

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

$$-n^{2}-n+90=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn+90. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-90=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -1.

$$a=9 b=-10$$

Rewrite -n^{2}-n+90 as \left(-n^{2}+9n\right)+\left(-10n+90\right).

$$\left(-n^{2}+9n\right)+\left(-10n+90\right)$$

Factor out n in the first and 10 in the second group.

$$n\left(-n+9\right)+10\left(-n+9\right)$$

Factor out common term -n+9 by using distributive property.

$$\left(-n+9\right)\left(n+10\right)$$

To find equation solutions, solve -n+9=0 and n+10=0.

$$n=9 n=-10$$

Substitute 9 for n in the equation \sqrt{90-n}=n.

$$\sqrt{90-9}=9$$

Simplify. The value n=9 satisfies the equation.

$$9=9$$

Substitute -10 for n in the equation \sqrt{90-n}=n.

$$\sqrt{90-\left(-10\right)}=-10$$

Simplify. The value n=-10 does not satisfy the equation because the left and the right hand side have opposite signs.

$$10=-10$$

Equation \sqrt{90-n}=n has a unique solution.

$$n=9$$