Solve sqrt{5-4x}-x=0

$$\sqrt { 5 - 4 x } - x = 0$$

Solve for x

$$x=1$$

Solution Steps

Subtract -x from both sides of the equation.

$$\sqrt{5-4x}=x$$

Square both sides of the equation.

$$\left(\sqrt{5-4x}\right)^{2}=x^{2}$$

Calculate \sqrt{5-4x} to the power of 2 and get 5-4x.

$$5-4x=x^{2}$$

Subtract x^{2} from both sides.

$$5-4x-x^{2}=0$$

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

$$-x^{2}-4x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-5=-5$$

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. The only such pair is the system solution.

$$a=1 b=-5$$

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

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

Factor out x in the first and 5 in the second group.

$$x\left(-x+1\right)+5\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+5\right)$$

To find equation solutions, solve -x+1=0 and x+5=0.

$$x=1 x=-5$$

Substitute 1 for x in the equation \sqrt{5-4x}-x=0.

$$\sqrt{5-4}-1=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

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

$$\sqrt{5-4\left(-5\right)}-\left(-5\right)=0$$

Simplify. The value x=-5 does not satisfy the equation.

$$10=0$$

Equation \sqrt{5-4x}=x has a unique solution.

$$x=1$$