Solve 2x-9=sqrt{5x+1}

$$2 x - 9 = \sqrt { 5 x + 1 }$$

Solve for x

$$x = \frac{\sqrt{401} + 41}{8} \approx 7.628123049$$

Solution Steps

Square both sides of the equation.

$$\left(2x-9\right)^{2}=\left(\sqrt{5x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-9\right)^{2}.

$$4x^{2}-36x+81=\left(\sqrt{5x+1}\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$4x^{2}-36x+81=5x+1$$

Subtract 5x from both sides.

$$4x^{2}-36x+81-5x=1$$

Combine -36x and -5x to get -41x.

$$4x^{2}-41x+81=1$$

Subtract 1 from both sides.

$$4x^{2}-41x+81-1=0$$

Subtract 1 from 81 to get 80.

$$4x^{2}-41x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -41 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 4\times 80}}{2\times 4}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 4\times 80}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-41\right)±\sqrt{1681-16\times 80}}{2\times 4}$$

Multiply -16 times 80.

$$x=\frac{-\left(-41\right)±\sqrt{1681-1280}}{2\times 4}$$

Add 1681 to -1280.

$$x=\frac{-\left(-41\right)±\sqrt{401}}{2\times 4}$$

The opposite of -41 is 41.

$$x=\frac{41±\sqrt{401}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{41±\sqrt{401}}{8}$$

Now solve the equation x=\frac{41±\sqrt{401}}{8} when ± is plus. Add 41 to \sqrt{401}.

$$x=\frac{\sqrt{401}+41}{8}$$

Now solve the equation x=\frac{41±\sqrt{401}}{8} when ± is minus. Subtract \sqrt{401} from 41.

$$x=\frac{41-\sqrt{401}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{401}+41}{8} x=\frac{41-\sqrt{401}}{8}$$

Substitute \frac{\sqrt{401}+41}{8} for x in the equation 2x-9=\sqrt{5x+1}.

$$2\times \frac{\sqrt{401}+41}{8}-9=\sqrt{5\times \frac{\sqrt{401}+41}{8}+1}$$

Simplify. The value x=\frac{\sqrt{401}+41}{8} satisfies the equation.

$$\frac{1}{4}\times 401^{\frac{1}{2}}+\frac{5}{4}=\frac{5}{4}+\frac{1}{4}\times 401^{\frac{1}{2}}$$

Substitute \frac{41-\sqrt{401}}{8} for x in the equation 2x-9=\sqrt{5x+1}.

$$2\times \frac{41-\sqrt{401}}{8}-9=\sqrt{5\times \frac{41-\sqrt{401}}{8}+1}$$

Simplify. The value x=\frac{41-\sqrt{401}}{8} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{5}{4}-\frac{1}{4}\times 401^{\frac{1}{2}}=-\left(\frac{5}{4}-\frac{1}{4}\times 401^{\frac{1}{2}}\right)$$

Equation 2x-9=\sqrt{5x+1} has a unique solution.

$$x=\frac{\sqrt{401}+41}{8}$$