Solve sqrt{2x+1}-sqrt{x+2}=2sqrt{3}

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

Solve for x

$$x=73$$

Solution Steps

Subtract -\sqrt{x+2} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$2x+1=4\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$2x+1=4\times 3+4\sqrt{3}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}$$

Multiply 4 and 3 to get 12.

$$2x+1=12+4\sqrt{3}\sqrt{x+2}+\left(\sqrt{x+2}\right)^{2}$$

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

$$2x+1=12+4\sqrt{3}\sqrt{x+2}+x+2$$

Add 12 and 2 to get 14.

$$2x+1=14+4\sqrt{3}\sqrt{x+2}+x$$

Subtract 14+x from both sides of the equation.

$$2x+1-\left(14+x\right)=4\sqrt{3}\sqrt{x+2}$$

To find the opposite of 14+x, find the opposite of each term.

$$2x+1-14-x=4\sqrt{3}\sqrt{x+2}$$

Subtract 14 from 1 to get -13.

$$2x-13-x=4\sqrt{3}\sqrt{x+2}$$

Combine 2x and -x to get x.

$$x-13=4\sqrt{3}\sqrt{x+2}$$

Square both sides of the equation.

$$\left(x-13\right)^{2}=\left(4\sqrt{3}\sqrt{x+2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-13\right)^{2}.

$$x^{2}-26x+169=\left(4\sqrt{3}\sqrt{x+2}\right)^{2}$$

Expand \left(4\sqrt{3}\sqrt{x+2}\right)^{2}.

$$x^{2}-26x+169=4^{2}\left(\sqrt{3}\right)^{2}\left(\sqrt{x+2}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$x^{2}-26x+169=16\left(\sqrt{3}\right)^{2}\left(\sqrt{x+2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$x^{2}-26x+169=16\times 3\left(\sqrt{x+2}\right)^{2}$$

Multiply 16 and 3 to get 48.

$$x^{2}-26x+169=48\left(\sqrt{x+2}\right)^{2}$$

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

$$x^{2}-26x+169=48\left(x+2\right)$$

Use the distributive property to multiply 48 by x+2.

$$x^{2}-26x+169=48x+96$$

Subtract 48x from both sides.

$$x^{2}-26x+169-48x=96$$

Combine -26x and -48x to get -74x.

$$x^{2}-74x+169=96$$

Subtract 96 from both sides.

$$x^{2}-74x+169-96=0$$

Subtract 96 from 169 to get 73.

$$x^{2}-74x+73=0$$

To solve the equation, factor x^{2}-74x+73 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-74 ab=73$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-73 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-73\right)\left(x-1\right)$$

To find equation solutions, solve x-73=0 and x-1=0.

$$x=73 x=1$$

Substitute 73 for x in the equation \sqrt{2x+1}-\sqrt{x+2}=2\sqrt{3}.

$$\sqrt{2\times 73+1}-\sqrt{73+2}=2\sqrt{3}$$

Simplify. The value x=73 satisfies the equation.

$$2\times 3^{\frac{1}{2}}=2\times 3^{\frac{1}{2}}$$

Substitute 1 for x in the equation \sqrt{2x+1}-\sqrt{x+2}=2\sqrt{3}.

$$\sqrt{2\times 1+1}-\sqrt{1+2}=2\sqrt{3}$$

Simplify. The value x=1 does not satisfy the equation.

$$0=2\times 3^{\frac{1}{2}}$$

Substitute 73 for x in the equation \sqrt{2x+1}-\sqrt{x+2}=2\sqrt{3}.

$$\sqrt{2\times 73+1}-\sqrt{73+2}=2\sqrt{3}$$

Simplify. The value x=73 satisfies the equation.

$$2\times 3^{\frac{1}{2}}=2\times 3^{\frac{1}{2}}$$

Equation \sqrt{2x+1}=\sqrt{x+2}+2\sqrt{3} has a unique solution.

$$x=73$$