Solve 110-4x=sqrt{2x+3}

$$110 - 4 x = \sqrt { 2 x + 3 }$$

Solve for x

$$x = \frac{441 - \sqrt{929}}{16} \approx 25.657531168$$

Solution Steps

Square both sides of the equation.

$$\left(110-4x\right)^{2}=\left(\sqrt{2x+3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(110-4x\right)^{2}.

$$12100-880x+16x^{2}=\left(\sqrt{2x+3}\right)^{2}$$

Calculate \sqrt{2x+3} to the power of 2 and get 2x+3.

$$12100-880x+16x^{2}=2x+3$$

Subtract 2x from both sides.

$$12100-880x+16x^{2}-2x=3$$

Combine -880x and -2x to get -882x.

$$12100-882x+16x^{2}=3$$

Subtract 3 from both sides.

$$12100-882x+16x^{2}-3=0$$

Subtract 3 from 12100 to get 12097.

$$12097-882x+16x^{2}=0$$

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.

$$16x^{2}-882x+12097=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -882 for b, and 12097 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-882\right)±\sqrt{\left(-882\right)^{2}-4\times 16\times 12097}}{2\times 16}$$

Square -882.

$$x=\frac{-\left(-882\right)±\sqrt{777924-4\times 16\times 12097}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-882\right)±\sqrt{777924-64\times 12097}}{2\times 16}$$

Multiply -64 times 12097.

$$x=\frac{-\left(-882\right)±\sqrt{777924-774208}}{2\times 16}$$

Add 777924 to -774208.

$$x=\frac{-\left(-882\right)±\sqrt{3716}}{2\times 16}$$

Take the square root of 3716.

$$x=\frac{-\left(-882\right)±2\sqrt{929}}{2\times 16}$$

The opposite of -882 is 882.

$$x=\frac{882±2\sqrt{929}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{882±2\sqrt{929}}{32}$$

Now solve the equation x=\frac{882±2\sqrt{929}}{32} when ± is plus. Add 882 to 2\sqrt{929}.

$$x=\frac{2\sqrt{929}+882}{32}$$

Divide 882+2\sqrt{929} by 32.

$$x=\frac{\sqrt{929}+441}{16}$$

Now solve the equation x=\frac{882±2\sqrt{929}}{32} when ± is minus. Subtract 2\sqrt{929} from 882.

$$x=\frac{882-2\sqrt{929}}{32}$$

Divide 882-2\sqrt{929} by 32.

$$x=\frac{441-\sqrt{929}}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{929}+441}{16} x=\frac{441-\sqrt{929}}{16}$$

Substitute \frac{\sqrt{929}+441}{16} for x in the equation 110-4x=\sqrt{2x+3}.

$$110-4\times \frac{\sqrt{929}+441}{16}=\sqrt{2\times \frac{\sqrt{929}+441}{16}+3}$$

Simplify. The value x=\frac{\sqrt{929}+441}{16} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{1}{4}-\frac{1}{4}\times 929^{\frac{1}{2}}=\frac{1}{4}+\frac{1}{4}\times 929^{\frac{1}{2}}$$

Substitute \frac{441-\sqrt{929}}{16} for x in the equation 110-4x=\sqrt{2x+3}.

$$110-4\times \frac{441-\sqrt{929}}{16}=\sqrt{2\times \frac{441-\sqrt{929}}{16}+3}$$

Simplify. The value x=\frac{441-\sqrt{929}}{16} satisfies the equation.

$$-\frac{1}{4}+\frac{1}{4}\times 929^{\frac{1}{2}}=-\left(\frac{1}{4}-\frac{1}{4}\times 929^{\frac{1}{2}}\right)$$

Equation 110-4x=\sqrt{2x+3} has a unique solution.

$$x=\frac{441-\sqrt{929}}{16}$$