Solve x=sqrt{1-5x}-1

$$x = \sqrt { 1 - 5 x } - 1$$

Solve for x

$$x=0$$

Solution Steps

Subtract -1 from both sides of the equation.

$$x+1=\sqrt{1-5x}$$

Square both sides of the equation.

$$\left(x+1\right)^{2}=\left(\sqrt{1-5x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1=\left(\sqrt{1-5x}\right)^{2}$$

Calculate \sqrt{1-5x} to the power of 2 and get 1-5x.

$$x^{2}+2x+1=1-5x$$

Subtract 1 from both sides.

$$x^{2}+2x+1-1=-5x$$

Subtract 1 from 1 to get 0.

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

Add 5x to both sides.

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

Combine 2x and 5x to get 7x.

$$x^{2}+7x=0$$

Factor out x.

$$x\left(x+7\right)=0$$

To find equation solutions, solve x=0 and x+7=0.

$$x=0 x=-7$$

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

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

Simplify. The value x=0 satisfies the equation.

$$0=0$$

Substitute -7 for x in the equation x=\sqrt{1-5x}-1.

$$-7=\sqrt{1-5\left(-7\right)}-1$$

Simplify. The value x=-7 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-7=5$$

Equation x+1=\sqrt{1-5x} has a unique solution.

$$x=0$$