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

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

Solve for x

$$x=1$$

Solution Steps

Subtract \sqrt{3x+1} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{3x+1}\right)^{2}.

$$x+3=16-8\sqrt{3x+1}+\left(\sqrt{3x+1}\right)^{2}$$

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

$$x+3=16-8\sqrt{3x+1}+3x+1$$

Add 16 and 1 to get 17.

$$x+3=17-8\sqrt{3x+1}+3x$$

Subtract 17+3x from both sides of the equation.

$$x+3-\left(17+3x\right)=-8\sqrt{3x+1}$$

To find the opposite of 17+3x, find the opposite of each term.

$$x+3-17-3x=-8\sqrt{3x+1}$$

Subtract 17 from 3 to get -14.

$$x-14-3x=-8\sqrt{3x+1}$$

Combine x and -3x to get -2x.

$$-2x-14=-8\sqrt{3x+1}$$

Square both sides of the equation.

$$\left(-2x-14\right)^{2}=\left(-8\sqrt{3x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-14\right)^{2}.

$$4x^{2}+56x+196=\left(-8\sqrt{3x+1}\right)^{2}$$

Expand \left(-8\sqrt{3x+1}\right)^{2}.

$$4x^{2}+56x+196=\left(-8\right)^{2}\left(\sqrt{3x+1}\right)^{2}$$

Calculate -8 to the power of 2 and get 64.

$$4x^{2}+56x+196=64\left(\sqrt{3x+1}\right)^{2}$$

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

$$4x^{2}+56x+196=64\left(3x+1\right)$$

Use the distributive property to multiply 64 by 3x+1.

$$4x^{2}+56x+196=192x+64$$

Subtract 192x from both sides.

$$4x^{2}+56x+196-192x=64$$

Combine 56x and -192x to get -136x.

$$4x^{2}-136x+196=64$$

Subtract 64 from both sides.

$$4x^{2}-136x+196-64=0$$

Subtract 64 from 196 to get 132.

$$4x^{2}-136x+132=0$$

Divide both sides by 4.

$$x^{2}-34x+33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+33. To find a and b, set up a system to be solved.

$$a+b=-34 ab=1\times 33=33$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 33.

$$-1,-33 -3,-11$$

Calculate the sum for each pair.

$$-1-33=-34 -3-11=-14$$

The solution is the pair that gives sum -34.

$$a=-33 b=-1$$

Rewrite x^{2}-34x+33 as \left(x^{2}-33x\right)+\left(-x+33\right).

$$\left(x^{2}-33x\right)+\left(-x+33\right)$$

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

$$x\left(x-33\right)-\left(x-33\right)$$

Factor out common term x-33 by using distributive property.

$$\left(x-33\right)\left(x-1\right)$$

To find equation solutions, solve x-33=0 and x-1=0.

$$x=33 x=1$$

Substitute 33 for x in the equation \sqrt{x+3}+\sqrt{3x+1}=4.

$$\sqrt{33+3}+\sqrt{3\times 33+1}=4$$

Simplify. The value x=33 does not satisfy the equation.

$$16=4$$

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

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

Simplify. The value x=1 satisfies the equation.

$$4=4$$

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

$$x=1$$