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

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

Solve for x

$$x=\frac{5\sqrt{33}-27}{2}\approx 0.861406616$$

Solution Steps

Subtract -1 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Expand \left(5\sqrt{1-x}\right)^{2}.

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

Calculate 5 to the power of 2 and get 25.

$$x^{2}+2x+1=25\left(\sqrt{1-x}\right)^{2}$$

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

$$x^{2}+2x+1=25\left(1-x\right)$$

Use the distributive property to multiply 25 by 1-x.

$$x^{2}+2x+1=25-25x$$

Subtract 25 from both sides.

$$x^{2}+2x+1-25=-25x$$

Subtract 25 from 1 to get -24.

$$x^{2}+2x-24=-25x$$

Add 25x to both sides.

$$x^{2}+2x-24+25x=0$$

Combine 2x and 25x to get 27x.

$$x^{2}+27x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 27 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-27±\sqrt{27^{2}-4\left(-24\right)}}{2}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$x=\frac{-27±\sqrt{729+96}}{2}$$

Add 729 to 96.

$$x=\frac{-27±\sqrt{825}}{2}$$

Take the square root of 825.

$$x=\frac{-27±5\sqrt{33}}{2}$$

Now solve the equation x=\frac{-27±5\sqrt{33}}{2} when ± is plus. Add -27 to 5\sqrt{33}.

$$x=\frac{5\sqrt{33}-27}{2}$$

Now solve the equation x=\frac{-27±5\sqrt{33}}{2} when ± is minus. Subtract 5\sqrt{33} from -27.

$$x=\frac{-5\sqrt{33}-27}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{33}-27}{2} x=\frac{-5\sqrt{33}-27}{2}$$

Substitute \frac{5\sqrt{33}-27}{2} for x in the equation x=5\sqrt{1-x}-1.

$$\frac{5\sqrt{33}-27}{2}=5\sqrt{1-\frac{5\sqrt{33}-27}{2}}-1$$

Simplify. The value x=\frac{5\sqrt{33}-27}{2} satisfies the equation.

$$\frac{5}{2}\times 33^{\frac{1}{2}}-\frac{27}{2}=-\frac{27}{2}+\frac{5}{2}\times 33^{\frac{1}{2}}$$

Substitute \frac{-5\sqrt{33}-27}{2} for x in the equation x=5\sqrt{1-x}-1.

$$\frac{-5\sqrt{33}-27}{2}=5\sqrt{1-\frac{-5\sqrt{33}-27}{2}}-1$$

Simplify. The value x=\frac{-5\sqrt{33}-27}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{5}{2}\times 33^{\frac{1}{2}}-\frac{27}{2}=\frac{23}{2}+\frac{5}{2}\times 33^{\frac{1}{2}}$$

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

$$x=\frac{5\sqrt{33}-27}{2}$$