Solve x-sqrt{6-5x}=0

$$x- \sqrt{ 6-5x } =0$$

Solve for x

$$x=1$$

Solution Steps

Subtract x from both sides of the equation.

$$-\sqrt{6-5x}=-x$$

Cancel out -1 on both sides.

$$\sqrt{6-5x}=x$$

Square both sides of the equation.

$$\left(\sqrt{6-5x}\right)^{2}=x^{2}$$

Calculate \sqrt{6-5x} to the power of 2 and get 6-5x.

$$6-5x=x^{2}$$

Subtract x^{2} from both sides.

$$6-5x-x^{2}=0$$

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

$$-x^{2}-5x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-6=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

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

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

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

$$x\left(-x+1\right)+6\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+6\right)$$

To find equation solutions, solve -x+1=0 and x+6=0.

$$x=1 x=-6$$

Substitute 1 for x in the equation x-\sqrt{6-5x}=0.

$$1-\sqrt{6-5}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -6 for x in the equation x-\sqrt{6-5x}=0.

$$-6-\sqrt{6-5\left(-6\right)}=0$$

Simplify. The value x=-6 does not satisfy the equation.

$$-12=0$$

Equation \sqrt{6-5x}=x has a unique solution.

$$x=1$$