Solve sqrt{30-x}=x

$$\sqrt { 30 - x } = x$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{30-x}\right)^{2}=x^{2}$$

Calculate \sqrt{30-x} to the power of 2 and get 30-x.

$$30-x=x^{2}$$

Subtract x^{2} from both sides.

$$30-x-x^{2}=0$$

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

$$-x^{2}-x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-30=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=5 b=-6$$

Rewrite -x^{2}-x+30 as \left(-x^{2}+5x\right)+\left(-6x+30\right).

$$\left(-x^{2}+5x\right)+\left(-6x+30\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(-x+5\right)+6\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(x+6\right)$$

To find equation solutions, solve -x+5=0 and x+6=0.

$$x=5 x=-6$$

Substitute 5 for x in the equation \sqrt{30-x}=x.

$$\sqrt{30-5}=5$$

Simplify. The value x=5 satisfies the equation.

$$5=5$$

Substitute -6 for x in the equation \sqrt{30-x}=x.

$$\sqrt{30-\left(-6\right)}=-6$$

Simplify. The value x=-6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$6=-6$$

Equation \sqrt{30-x}=x has a unique solution.

$$x=5$$