Solve sqrt{3x-11}+sqrt{3x}=sqrt{12x-23}

$$\sqrt{ 3x-11 } + \sqrt{ 3x } = \sqrt{ 12x-23 }$$

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x-11}+\sqrt{3x}\right)^{2}=\left(\sqrt{12x-23}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3x-11}+\sqrt{3x}\right)^{2}.

$$\left(\sqrt{3x-11}\right)^{2}+2\sqrt{3x-11}\sqrt{3x}+\left(\sqrt{3x}\right)^{2}=\left(\sqrt{12x-23}\right)^{2}$$

Calculate \sqrt{3x-11} to the power of 2 and get 3x-11.

$$3x-11+2\sqrt{3x-11}\sqrt{3x}+\left(\sqrt{3x}\right)^{2}=\left(\sqrt{12x-23}\right)^{2}$$

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

$$3x-11+2\sqrt{3x-11}\sqrt{3x}+3x=\left(\sqrt{12x-23}\right)^{2}$$

Combine 3x and 3x to get 6x.

$$6x-11+2\sqrt{3x-11}\sqrt{3x}=\left(\sqrt{12x-23}\right)^{2}$$

Calculate \sqrt{12x-23} to the power of 2 and get 12x-23.

$$6x-11+2\sqrt{3x-11}\sqrt{3x}=12x-23$$

Subtract 6x-11 from both sides of the equation.

$$2\sqrt{3x-11}\sqrt{3x}=12x-23-\left(6x-11\right)$$

To find the opposite of 6x-11, find the opposite of each term.

$$2\sqrt{3x-11}\sqrt{3x}=12x-23-6x+11$$

Combine 12x and -6x to get 6x.

$$2\sqrt{3x-11}\sqrt{3x}=6x-23+11$$

Add -23 and 11 to get -12.

$$2\sqrt{3x-11}\sqrt{3x}=6x-12$$

Square both sides of the equation.

$$\left(2\sqrt{3x-11}\sqrt{3x}\right)^{2}=\left(6x-12\right)^{2}$$

Expand \left(2\sqrt{3x-11}\sqrt{3x}\right)^{2}.

$$2^{2}\left(\sqrt{3x-11}\right)^{2}\left(\sqrt{3x}\right)^{2}=\left(6x-12\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3x-11}\right)^{2}\left(\sqrt{3x}\right)^{2}=\left(6x-12\right)^{2}$$

Calculate \sqrt{3x-11} to the power of 2 and get 3x-11.

$$4\left(3x-11\right)\left(\sqrt{3x}\right)^{2}=\left(6x-12\right)^{2}$$

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

$$4\left(3x-11\right)\times 3x=\left(6x-12\right)^{2}$$

Multiply 4 and 3 to get 12.

$$12\left(3x-11\right)x=\left(6x-12\right)^{2}$$

Use the distributive property to multiply 12 by 3x-11.

$$\left(36x-132\right)x=\left(6x-12\right)^{2}$$

Use the distributive property to multiply 36x-132 by x.

$$36x^{2}-132x=\left(6x-12\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-12\right)^{2}.

$$36x^{2}-132x=36x^{2}-144x+144$$

Subtract 36x^{2} from both sides.

$$36x^{2}-132x-36x^{2}=-144x+144$$

Combine 36x^{2} and -36x^{2} to get 0.

$$-132x=-144x+144$$

Add 144x to both sides.

$$-132x+144x=144$$

Combine -132x and 144x to get 12x.

$$12x=144$$

Divide both sides by 12.

$$x=\frac{144}{12}$$

Divide 144 by 12 to get 12.

$$x=12$$

Substitute 12 for x in the equation \sqrt{3x-11}+\sqrt{3x}=\sqrt{12x-23}.

$$\sqrt{3\times 12-11}+\sqrt{3\times 12}=\sqrt{12\times 12-23}$$

Simplify. The value x=12 satisfies the equation.

$$11=11$$

Equation \sqrt{3x-11}+\sqrt{3x}=\sqrt{12x-23} has a unique solution.

$$x=12$$