Solve 3=sqrt{7*x}+1/8x

$$3 = \sqrt { 7 \times x } + \frac { 1 } { 8 } x$$

Solve for x

$$x=248-16\sqrt{238}\approx 1.164022071$$

Solution Steps

Subtract \frac{1}{8}x from both sides of the equation.

$$3-\frac{1}{8}x=\sqrt{7x}$$

Square both sides of the equation.

$$\left(3-\frac{1}{8}x\right)^{2}=\left(\sqrt{7x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\frac{1}{8}x\right)^{2}.

$$9-\frac{3}{4}x+\frac{1}{64}x^{2}=\left(\sqrt{7x}\right)^{2}$$

Calculate \sqrt{7x} to the power of 2 and get 7x.

$$9-\frac{3}{4}x+\frac{1}{64}x^{2}=7x$$

Subtract 7x from both sides.

$$9-\frac{3}{4}x+\frac{1}{64}x^{2}-7x=0$$

Combine -\frac{3}{4}x and -7x to get -\frac{31}{4}x.

$$9-\frac{31}{4}x+\frac{1}{64}x^{2}=0$$

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.

$$\frac{1}{64}x^{2}-\frac{31}{4}x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{64} for a, -\frac{31}{4} for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{31}{4}\right)±\sqrt{\left(-\frac{31}{4}\right)^{2}-4\times \frac{1}{64}\times 9}}{2\times \frac{1}{64}}$$

Square -\frac{31}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{31}{4}\right)±\sqrt{\frac{961}{16}-4\times \frac{1}{64}\times 9}}{2\times \frac{1}{64}}$$

Multiply -4 times \frac{1}{64}.

$$x=\frac{-\left(-\frac{31}{4}\right)±\sqrt{\frac{961}{16}-\frac{1}{16}\times 9}}{2\times \frac{1}{64}}$$

Multiply -\frac{1}{16} times 9.

$$x=\frac{-\left(-\frac{31}{4}\right)±\sqrt{\frac{961-9}{16}}}{2\times \frac{1}{64}}$$

Add \frac{961}{16} to -\frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{31}{4}\right)±\sqrt{\frac{119}{2}}}{2\times \frac{1}{64}}$$

Take the square root of \frac{119}{2}.

$$x=\frac{-\left(-\frac{31}{4}\right)±\frac{\sqrt{238}}{2}}{2\times \frac{1}{64}}$$

The opposite of -\frac{31}{4} is \frac{31}{4}.

$$x=\frac{\frac{31}{4}±\frac{\sqrt{238}}{2}}{2\times \frac{1}{64}}$$

Multiply 2 times \frac{1}{64}.

$$x=\frac{\frac{31}{4}±\frac{\sqrt{238}}{2}}{\frac{1}{32}}$$

Now solve the equation x=\frac{\frac{31}{4}±\frac{\sqrt{238}}{2}}{\frac{1}{32}} when ± is plus. Add \frac{31}{4} to \frac{\sqrt{238}}{2}.

$$x=\frac{\frac{\sqrt{238}}{2}+\frac{31}{4}}{\frac{1}{32}}$$

Divide \frac{31}{4}+\frac{\sqrt{238}}{2} by \frac{1}{32} by multiplying \frac{31}{4}+\frac{\sqrt{238}}{2} by the reciprocal of \frac{1}{32}.

$$x=16\sqrt{238}+248$$

Now solve the equation x=\frac{\frac{31}{4}±\frac{\sqrt{238}}{2}}{\frac{1}{32}} when ± is minus. Subtract \frac{\sqrt{238}}{2} from \frac{31}{4}.

$$x=\frac{-\frac{\sqrt{238}}{2}+\frac{31}{4}}{\frac{1}{32}}$$

Divide \frac{31}{4}-\frac{\sqrt{238}}{2} by \frac{1}{32} by multiplying \frac{31}{4}-\frac{\sqrt{238}}{2} by the reciprocal of \frac{1}{32}.

$$x=248-16\sqrt{238}$$

The equation is now solved.

$$x=16\sqrt{238}+248 x=248-16\sqrt{238}$$

Substitute 16\sqrt{238}+248 for x in the equation 3=\sqrt{7x}+\frac{1}{8}x.

$$3=\sqrt{7\left(16\sqrt{238}+248\right)}+\frac{1}{8}\left(16\sqrt{238}+248\right)$$

Simplify. The value x=16\sqrt{238}+248 does not satisfy the equation.

$$3=4\times 238^{\frac{1}{2}}+59$$

Substitute 248-16\sqrt{238} for x in the equation 3=\sqrt{7x}+\frac{1}{8}x.

$$3=\sqrt{7\left(248-16\sqrt{238}\right)}+\frac{1}{8}\left(248-16\sqrt{238}\right)$$

Simplify. The value x=248-16\sqrt{238} satisfies the equation.

$$3=3$$

Equation -\frac{x}{8}+3=\sqrt{7x} has a unique solution.

$$x=248-16\sqrt{238}$$