Solve 1+8x=sqrt{x}

$$1+8x= \sqrt{ x }$$

Solve for x (complex solution)

$$x=\frac{-15+\sqrt{31}i}{128}\approx -0.1171875+0.043498159i$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+8x\right)^{2}.

$$1+16x+64x^{2}=\left(\sqrt{x}\right)^{2}$$

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

$$1+16x+64x^{2}=x$$

Subtract x from both sides.

$$1+16x+64x^{2}-x=0$$

Combine 16x and -x to get 15x.

$$1+15x+64x^{2}=0$$

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.

$$64x^{2}+15x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 64 for a, 15 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\times 64}}{2\times 64}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 64}}{2\times 64}$$

Multiply -4 times 64.

$$x=\frac{-15±\sqrt{225-256}}{2\times 64}$$

Add 225 to -256.

$$x=\frac{-15±\sqrt{-31}}{2\times 64}$$

Take the square root of -31.

$$x=\frac{-15±\sqrt{31}i}{2\times 64}$$

Multiply 2 times 64.

$$x=\frac{-15±\sqrt{31}i}{128}$$

Now solve the equation x=\frac{-15±\sqrt{31}i}{128} when ± is plus. Add -15 to i\sqrt{31}.

$$x=\frac{-15+\sqrt{31}i}{128}$$

Now solve the equation x=\frac{-15±\sqrt{31}i}{128} when ± is minus. Subtract i\sqrt{31} from -15.

$$x=\frac{-\sqrt{31}i-15}{128}$$

The equation is now solved.

$$x=\frac{-15+\sqrt{31}i}{128} x=\frac{-\sqrt{31}i-15}{128}$$

Substitute \frac{-15+\sqrt{31}i}{128} for x in the equation 1+8x=\sqrt{x}.

$$1+8\times \frac{-15+\sqrt{31}i}{128}=\sqrt{\frac{-15+\sqrt{31}i}{128}}$$

Simplify. The value x=\frac{-15+\sqrt{31}i}{128} satisfies the equation.

$$\frac{1}{16}+\frac{1}{16}i\times 31^{\frac{1}{2}}=\frac{1}{16}+\frac{1}{16}i\times 31^{\frac{1}{2}}$$

Substitute \frac{-\sqrt{31}i-15}{128} for x in the equation 1+8x=\sqrt{x}.

$$1+8\times \frac{-\sqrt{31}i-15}{128}=\sqrt{\frac{-\sqrt{31}i-15}{128}}$$

Simplify. The value x=\frac{-\sqrt{31}i-15}{128} does not satisfy the equation.

$$\frac{1}{16}-\frac{1}{16}i\times 31^{\frac{1}{2}}=-\left(\frac{1}{16}-\frac{1}{16}i\times 31^{\frac{1}{2}}\right)$$

Equation 8x+1=\sqrt{x} has a unique solution.

$$x=\frac{-15+\sqrt{31}i}{128}$$