Solve sqrt{-x+12}=x

$$\sqrt { - x + 12 } = x$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

$$-x+12=x^{2}$$

Subtract x^{2} from both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+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 -x^{2}-x+12 as \left(-x^{2}+3x\right)+\left(-4x+12\right).

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

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

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

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

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

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

$$x=3 x=-4$$

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

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

Simplify. The value x=3 satisfies the equation.

$$3=3$$

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

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

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

$$4=-4$$

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

$$x=3$$