Solve sqrt{1div2+1div4+1div8+1div16+1div2x}=x

$$\sqrt{ 1 \div 2+1 \div 4+1 \div 8+1 \div 16+1 \div 2x } =x$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Least common multiple of 2 and 4 is 4. Convert \frac{1}{2} and \frac{1}{4} to fractions with denominator 4.

$$\left(\sqrt{\frac{2}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Since \frac{2}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$\left(\sqrt{\frac{2+1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Add 2 and 1 to get 3.

$$\left(\sqrt{\frac{3}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Least common multiple of 4 and 8 is 8. Convert \frac{3}{4} and \frac{1}{8} to fractions with denominator 8.

$$\left(\sqrt{\frac{6}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Since \frac{6}{8} and \frac{1}{8} have the same denominator, add them by adding their numerators.

$$\left(\sqrt{\frac{6+1}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Add 6 and 1 to get 7.

$$\left(\sqrt{\frac{7}{8}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Least common multiple of 8 and 16 is 16. Convert \frac{7}{8} and \frac{1}{16} to fractions with denominator 16.

$$\left(\sqrt{\frac{14}{16}+\frac{1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Since \frac{14}{16} and \frac{1}{16} have the same denominator, add them by adding their numerators.

$$\left(\sqrt{\frac{14+1}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Add 14 and 1 to get 15.

$$\left(\sqrt{\frac{15}{16}+\frac{1}{2}x}\right)^{2}=x^{2}$$

Calculate \sqrt{\frac{15}{16}+\frac{1}{2}x} to the power of 2 and get \frac{15}{16}+\frac{1}{2}x.

$$\frac{15}{16}+\frac{1}{2}x=x^{2}$$

Subtract x^{2} from both sides.

$$\frac{15}{16}+\frac{1}{2}x-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+\frac{1}{2}x+\frac{15}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, \frac{1}{2} for b, and \frac{15}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-1\right)\times \frac{15}{16}}}{2\left(-1\right)}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-1\right)\times \frac{15}{16}}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+4\times \frac{15}{16}}}{2\left(-1\right)}$$

Multiply 4 times \frac{15}{16}.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1+15}{4}}}{2\left(-1\right)}$$

Add \frac{1}{4} to \frac{15}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1}{2}±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$x=\frac{-\frac{1}{2}±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-\frac{1}{2}±2}{-2}$$

Now solve the equation x=\frac{-\frac{1}{2}±2}{-2} when ± is plus. Add -\frac{1}{2} to 2.

$$x=\frac{\frac{3}{2}}{-2}$$

Divide \frac{3}{2} by -2.

$$x=-\frac{3}{4}$$

Now solve the equation x=\frac{-\frac{1}{2}±2}{-2} when ± is minus. Subtract 2 from -\frac{1}{2}.

$$x=-\frac{\frac{5}{2}}{-2}$$

Divide -\frac{5}{2} by -2.

$$x=\frac{5}{4}$$

The equation is now solved.

$$x=-\frac{3}{4} x=\frac{5}{4}$$

Substitute -\frac{3}{4} for x in the equation \sqrt{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}=x.

$$\sqrt{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}\left(-\frac{3}{4}\right)}=-\frac{3}{4}$$

Simplify. The value x=-\frac{3}{4} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{3}{4}=-\frac{3}{4}$$

Substitute \frac{5}{4} for x in the equation \sqrt{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}x}=x.

$$\sqrt{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{2}\times \frac{5}{4}}=\frac{5}{4}$$

Simplify. The value x=\frac{5}{4} satisfies the equation.

$$\frac{5}{4}=\frac{5}{4}$$

Equation \sqrt{\frac{x}{2}+\frac{15}{16}}=x has a unique solution.

$$x=\frac{5}{4}$$