Solve sqrt{x+3}=sqrt{5x}+1

$$\sqrt { x + 3 } = \sqrt { 5 x } + 1$$

Solve for x

$$x=\frac{9-\sqrt{65}}{8}\approx 0.117217781$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+3}\right)^{2}=\left(\sqrt{5x}+1\right)^{2}$$

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

$$x+3=\left(\sqrt{5x}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5x}+1\right)^{2}.

$$x+3=\left(\sqrt{5x}\right)^{2}+2\sqrt{5x}+1$$

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

$$x+3=5x+2\sqrt{5x}+1$$

Subtract 5x+1 from both sides of the equation.

$$x+3-\left(5x+1\right)=2\sqrt{5x}$$

To find the opposite of 5x+1, find the opposite of each term.

$$x+3-5x-1=2\sqrt{5x}$$

Combine x and -5x to get -4x.

$$-4x+3-1=2\sqrt{5x}$$

Subtract 1 from 3 to get 2.

$$-4x+2=2\sqrt{5x}$$

Square both sides of the equation.

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

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

$$16x^{2}-16x+4=\left(2\sqrt{5x}\right)^{2}$$

Expand \left(2\sqrt{5x}\right)^{2}.

$$16x^{2}-16x+4=2^{2}\left(\sqrt{5x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$16x^{2}-16x+4=4\left(\sqrt{5x}\right)^{2}$$

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

$$16x^{2}-16x+4=4\times 5x$$

Multiply 4 and 5 to get 20.

$$16x^{2}-16x+4=20x$$

Subtract 20x from both sides.

$$16x^{2}-16x+4-20x=0$$

Combine -16x and -20x to get -36x.

$$16x^{2}-36x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -36 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 16\times 4}}{2\times 16}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 16\times 4}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-36\right)±\sqrt{1296-64\times 4}}{2\times 16}$$

Multiply -64 times 4.

$$x=\frac{-\left(-36\right)±\sqrt{1296-256}}{2\times 16}$$

Add 1296 to -256.

$$x=\frac{-\left(-36\right)±\sqrt{1040}}{2\times 16}$$

Take the square root of 1040.

$$x=\frac{-\left(-36\right)±4\sqrt{65}}{2\times 16}$$

The opposite of -36 is 36.

$$x=\frac{36±4\sqrt{65}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{36±4\sqrt{65}}{32}$$

Now solve the equation x=\frac{36±4\sqrt{65}}{32} when ± is plus. Add 36 to 4\sqrt{65}.

$$x=\frac{4\sqrt{65}+36}{32}$$

Divide 36+4\sqrt{65} by 32.

$$x=\frac{\sqrt{65}+9}{8}$$

Now solve the equation x=\frac{36±4\sqrt{65}}{32} when ± is minus. Subtract 4\sqrt{65} from 36.

$$x=\frac{36-4\sqrt{65}}{32}$$

Divide 36-4\sqrt{65} by 32.

$$x=\frac{9-\sqrt{65}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{65}+9}{8} x=\frac{9-\sqrt{65}}{8}$$

Substitute \frac{\sqrt{65}+9}{8} for x in the equation \sqrt{x+3}=\sqrt{5x}+1.

$$\sqrt{\frac{\sqrt{65}+9}{8}+3}=\sqrt{5\times \frac{\sqrt{65}+9}{8}}+1$$

Simplify. The value x=\frac{\sqrt{65}+9}{8} does not satisfy the equation.

$$\frac{1}{4}+\frac{1}{4}\times 65^{\frac{1}{2}}=\frac{9}{4}+\frac{1}{4}\times 65^{\frac{1}{2}}$$

Substitute \frac{9-\sqrt{65}}{8} for x in the equation \sqrt{x+3}=\sqrt{5x}+1.

$$\sqrt{\frac{9-\sqrt{65}}{8}+3}=\sqrt{5\times \frac{9-\sqrt{65}}{8}}+1$$

Simplify. The value x=\frac{9-\sqrt{65}}{8} satisfies the equation.

$$-\left(\frac{1}{4}-\frac{1}{4}\times 65^{\frac{1}{2}}\right)=-\frac{1}{4}+\frac{1}{4}\times 65^{\frac{1}{2}}$$

Equation \sqrt{x+3}=\sqrt{5x}+1 has a unique solution.

$$x=\frac{9-\sqrt{65}}{8}$$