Solve sqrt{0.4x}+sqrt{0.36+0.4x}=1

$$\sqrt{ 0.4x } + \sqrt{ 0.36+0.4x } =1$$

Solve for x

$$x=0.256$$

Solution Steps

Subtract \sqrt{0.36+0.4x} from both sides of the equation.

$$\sqrt{0.4x}=1-\sqrt{0.36+0.4x}$$

Square both sides of the equation.

$$\left(\sqrt{0.4x}\right)^{2}=\left(1-\sqrt{0.36+0.4x}\right)^{2}$$

Calculate \sqrt{0.4x} to the power of 2 and get 0.4x.

$$0.4x=\left(1-\sqrt{0.36+0.4x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{0.36+0.4x}\right)^{2}.

$$0.4x=1-2\sqrt{0.36+0.4x}+\left(\sqrt{0.36+0.4x}\right)^{2}$$

Calculate \sqrt{0.36+0.4x} to the power of 2 and get 0.36+0.4x.

$$0.4x=1-2\sqrt{0.36+0.4x}+0.36+0.4x$$

Add 1 and 0.36 to get 1.36.

$$0.4x=1.36-2\sqrt{0.36+0.4x}+0.4x$$

Add 2\sqrt{0.36+0.4x} to both sides.

$$0.4x+2\sqrt{0.36+0.4x}=1.36+0.4x$$

Subtract 0.4x from both sides.

$$0.4x+2\sqrt{0.36+0.4x}-0.4x=1.36$$

Combine 0.4x and -0.4x to get 0.

$$2\sqrt{0.36+0.4x}=1.36$$

Divide both sides by 2.

$$\sqrt{0.36+0.4x}=\frac{1.36}{2}$$

Expand \frac{1.36}{2} by multiplying both numerator and the denominator by 100.

$$\sqrt{0.36+0.4x}=\frac{136}{200}$$

Reduce the fraction \frac{136}{200} to lowest terms by extracting and canceling out 8.

$$\sqrt{0.36+0.4x}=\frac{17}{25}$$

Square both sides of the equation.

$$0.4x+0.36=\frac{289}{625}$$

Subtract 0.36 from both sides of the equation.

$$0.4x+0.36-0.36=\frac{289}{625}-0.36$$

Subtracting 0.36 from itself leaves 0.

$$0.4x=\frac{289}{625}-0.36$$

Subtract 0.36 from \frac{289}{625} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$0.4x=\frac{64}{625}$$

Divide both sides of the equation by 0.4, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{0.4x}{0.4}=\frac{\frac{64}{625}}{0.4}$$

Dividing by 0.4 undoes the multiplication by 0.4.

$$x=\frac{\frac{64}{625}}{0.4}$$

Divide \frac{64}{625} by 0.4 by multiplying \frac{64}{625} by the reciprocal of 0.4.

$$x=\frac{32}{125}$$

Substitute \frac{32}{125} for x in the equation \sqrt{0.4x}+\sqrt{0.36+0.4x}=1.

$$\sqrt{0.4\times \frac{32}{125}}+\sqrt{0.36+0.4\times \frac{32}{125}}=1$$

Simplify. The value x=\frac{32}{125} satisfies the equation.

$$1=1$$

Equation \sqrt{\frac{2x}{5}}=-\sqrt{\frac{2x}{5}+0.36}+1 has a unique solution.

$$x=\frac{32}{125}$$