Solve 4+sqrt{3x+1}=5+sqrt{x+4}

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

Solve for x

$$x=5$$

Solution Steps

Subtract 4 from both sides of the equation.

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

Subtract 4 from 5 to get 1.

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

Square both sides of the equation.

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

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{x+4}\right)^{2}.

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

Calculate \sqrt{x+4} to the power of 2 and get x+4.

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

Add 1 and 4 to get 5.

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

Subtract 5+x from both sides of the equation.

$$3x+1-\left(5+x\right)=2\sqrt{x+4}$$

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

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

Subtract 5 from 1 to get -4.

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

Combine 3x and -x to get 2x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{x+4} to the power of 2 and get x+4.

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

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

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

Subtract 4x from both sides.

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

Combine -16x and -4x to get -20x.

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

Subtract 16 from both sides.

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

Subtract 16 from 16 to get 0.

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

Factor out x.

$$x\left(4x-20\right)=0$$

To find equation solutions, solve x=0 and 4x-20=0.

$$x=0 x=5$$

Substitute 0 for x in the equation 4+\sqrt{3x+1}=5+\sqrt{x+4}.

$$4+\sqrt{3\times 0+1}=5+\sqrt{0+4}$$

Simplify. The value x=0 does not satisfy the equation.

$$5=7$$

Substitute 5 for x in the equation 4+\sqrt{3x+1}=5+\sqrt{x+4}.

$$4+\sqrt{3\times 5+1}=5+\sqrt{5+4}$$

Simplify. The value x=5 satisfies the equation.

$$8=8$$

Equation \sqrt{3x+1}=\sqrt{x+4}+1 has a unique solution.

$$x=5$$