Solve x-sqrt{36-5x}=0

$$x- \sqrt{ 36-5x } = 0$$

Solve for x

$$x=4$$

Solution Steps

Subtract x from both sides of the equation.

$$-\sqrt{36-5x}=-x$$

Cancel out -1 on both sides.

$$\sqrt{36-5x}=x$$

Square both sides of the equation.

$$\left(\sqrt{36-5x}\right)^{2}=x^{2}$$

Calculate \sqrt{36-5x} to the power of 2 and get 36-5x.

$$36-5x=x^{2}$$

Subtract x^{2} from both sides.

$$36-5x-x^{2}=0$$

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

$$-x^{2}-5x+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-36=-36$$

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 -36.

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

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=4 b=-9$$

Rewrite -x^{2}-5x+36 as \left(-x^{2}+4x\right)+\left(-9x+36\right).

$$\left(-x^{2}+4x\right)+\left(-9x+36\right)$$

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

$$x\left(-x+4\right)+9\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(x+9\right)$$

To find equation solutions, solve -x+4=0 and x+9=0.

$$x=4 x=-9$$

Substitute 4 for x in the equation x-\sqrt{36-5x}=0.

$$4-\sqrt{36-5\times 4}=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Substitute -9 for x in the equation x-\sqrt{36-5x}=0.

$$-9-\sqrt{36-5\left(-9\right)}=0$$

Simplify. The value x=-9 does not satisfy the equation.

$$-18=0$$

Equation \sqrt{36-5x}=x has a unique solution.

$$x=4$$