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

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add -4 and 4 to get 0.

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

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

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

Subtract 4x from both sides of the equation.

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

Combine 3x and -4x to get -x.

$$8\sqrt{x-1}=-x+1$$

Square both sides of the equation.

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

Expand \left(8\sqrt{x-1}\right)^{2}.

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

Calculate 8 to the power of 2 and get 64.

$$64\left(\sqrt{x-1}\right)^{2}=\left(-x+1\right)^{2}$$

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

$$64\left(x-1\right)=\left(-x+1\right)^{2}$$

Use the distributive property to multiply 64 by x-1.

$$64x-64=\left(-x+1\right)^{2}$$

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

$$64x-64=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$64x-64-x^{2}=-2x+1$$

Add 2x to both sides.

$$64x-64-x^{2}+2x=1$$

Combine 64x and 2x to get 66x.

$$66x-64-x^{2}=1$$

Subtract 1 from both sides.

$$66x-64-x^{2}-1=0$$

Subtract 1 from -64 to get -65.

$$66x-65-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+66x-65=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-65. To find a and b, set up a system to be solved.

$$a+b=66 ab=-\left(-65\right)=65$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 65.

$$1,65 5,13$$

Calculate the sum for each pair.

$$1+65=66 5+13=18$$

The solution is the pair that gives sum 66.

$$a=65 b=1$$

Rewrite -x^{2}+66x-65 as \left(-x^{2}+65x\right)+\left(x-65\right).

$$\left(-x^{2}+65x\right)+\left(x-65\right)$$

Factor out -x in -x^{2}+65x.

$$-x\left(x-65\right)+x-65$$

Factor out common term x-65 by using distributive property.

$$\left(x-65\right)\left(-x+1\right)$$

To find equation solutions, solve x-65=0 and -x+1=0.

$$x=65 x=1$$

Substitute 65 for x in the equation 2\sqrt{x-1}+2=\sqrt{3x+1}.

$$2\sqrt{65-1}+2=\sqrt{3\times 65+1}$$

Simplify. The value x=65 does not satisfy the equation.

$$18=14$$

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

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

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

$$x=1$$