Solve sqrt{x+1}+sqrt{3x+7}=2sqrt{2x+3}

$$\sqrt { x + 1 } + \sqrt { 3 x + 7 } = 2 \sqrt { 2 x + 3 }$$

Solve for x

$$x=3$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+1}+\sqrt{3x+7}\right)^{2}=\left(2\sqrt{2x+3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+1}+\sqrt{3x+7}\right)^{2}.

$$\left(\sqrt{x+1}\right)^{2}+2\sqrt{x+1}\sqrt{3x+7}+\left(\sqrt{3x+7}\right)^{2}=\left(2\sqrt{2x+3}\right)^{2}$$

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

$$x+1+2\sqrt{x+1}\sqrt{3x+7}+\left(\sqrt{3x+7}\right)^{2}=\left(2\sqrt{2x+3}\right)^{2}$$

Calculate \sqrt{3x+7} to the power of 2 and get 3x+7.

$$x+1+2\sqrt{x+1}\sqrt{3x+7}+3x+7=\left(2\sqrt{2x+3}\right)^{2}$$

Combine x and 3x to get 4x.

$$4x+1+2\sqrt{x+1}\sqrt{3x+7}+7=\left(2\sqrt{2x+3}\right)^{2}$$

Add 1 and 7 to get 8.

$$4x+8+2\sqrt{x+1}\sqrt{3x+7}=\left(2\sqrt{2x+3}\right)^{2}$$

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

$$4x+8+2\sqrt{x+1}\sqrt{3x+7}=2^{2}\left(\sqrt{2x+3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x+8+2\sqrt{x+1}\sqrt{3x+7}=4\left(\sqrt{2x+3}\right)^{2}$$

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

$$4x+8+2\sqrt{x+1}\sqrt{3x+7}=4\left(2x+3\right)$$

Use the distributive property to multiply 4 by 2x+3.

$$4x+8+2\sqrt{x+1}\sqrt{3x+7}=8x+12$$

Subtract 4x+8 from both sides of the equation.

$$2\sqrt{x+1}\sqrt{3x+7}=8x+12-\left(4x+8\right)$$

To find the opposite of 4x+8, find the opposite of each term.

$$2\sqrt{x+1}\sqrt{3x+7}=8x+12-4x-8$$

Combine 8x and -4x to get 4x.

$$2\sqrt{x+1}\sqrt{3x+7}=4x+12-8$$

Subtract 8 from 12 to get 4.

$$2\sqrt{x+1}\sqrt{3x+7}=4x+4$$

Square both sides of the equation.

$$\left(2\sqrt{x+1}\sqrt{3x+7}\right)^{2}=\left(4x+4\right)^{2}$$

Expand \left(2\sqrt{x+1}\sqrt{3x+7}\right)^{2}.

$$2^{2}\left(\sqrt{x+1}\right)^{2}\left(\sqrt{3x+7}\right)^{2}=\left(4x+4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+1}\right)^{2}\left(\sqrt{3x+7}\right)^{2}=\left(4x+4\right)^{2}$$

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

$$4\left(x+1\right)\left(\sqrt{3x+7}\right)^{2}=\left(4x+4\right)^{2}$$

Calculate \sqrt{3x+7} to the power of 2 and get 3x+7.

$$4\left(x+1\right)\left(3x+7\right)=\left(4x+4\right)^{2}$$

Use the distributive property to multiply 4 by x+1.

$$\left(4x+4\right)\left(3x+7\right)=\left(4x+4\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+4 by each term of 3x+7.

$$12x^{2}+28x+12x+28=\left(4x+4\right)^{2}$$

Combine 28x and 12x to get 40x.

$$12x^{2}+40x+28=\left(4x+4\right)^{2}$$

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

$$12x^{2}+40x+28=16x^{2}+32x+16$$

Subtract 16x^{2} from both sides.

$$12x^{2}+40x+28-16x^{2}=32x+16$$

Combine 12x^{2} and -16x^{2} to get -4x^{2}.

$$-4x^{2}+40x+28=32x+16$$

Subtract 32x from both sides.

$$-4x^{2}+40x+28-32x=16$$

Combine 40x and -32x to get 8x.

$$-4x^{2}+8x+28=16$$

Subtract 16 from both sides.

$$-4x^{2}+8x+28-16=0$$

Subtract 16 from 28 to get 12.

$$-4x^{2}+8x+12=0$$

Divide both sides by 4.

$$-x^{2}+2x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -x^{2}+2x+3 as \left(-x^{2}+3x\right)+\left(-x+3\right).

$$\left(-x^{2}+3x\right)+\left(-x+3\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x-1\right)$$

To find equation solutions, solve x-3=0 and -x-1=0.

$$x=3 x=-1$$

Substitute 3 for x in the equation \sqrt{x+1}+\sqrt{3x+7}=2\sqrt{2x+3}.

$$\sqrt{3+1}+\sqrt{3\times 3+7}=2\sqrt{2\times 3+3}$$

Simplify. The value x=3 satisfies the equation.

$$6=6$$

Substitute -1 for x in the equation \sqrt{x+1}+\sqrt{3x+7}=2\sqrt{2x+3}.

$$\sqrt{-1+1}+\sqrt{3\left(-1\right)+7}=2\sqrt{2\left(-1\right)+3}$$

Simplify. The value x=-1 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{x+1}+\sqrt{3x+7}=2\sqrt{2x+3}.

$$x=3 x=-1$$