Solve sqrt{6x+7}-sqrt{3x+3}=1

$$\sqrt{ 6x+7 } - \sqrt{ 3x+3 } =1$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-1$$

Solution Steps

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

$$\sqrt{6x+7}=1+\sqrt{3x+3}$$

Square both sides of the equation.

$$\left(\sqrt{6x+7}\right)^{2}=\left(1+\sqrt{3x+3}\right)^{2}$$

Calculate \sqrt{6x+7} to the power of 2 and get 6x+7.

$$6x+7=\left(1+\sqrt{3x+3}\right)^{2}$$

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

$$6x+7=1+2\sqrt{3x+3}+\left(\sqrt{3x+3}\right)^{2}$$

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

$$6x+7=1+2\sqrt{3x+3}+3x+3$$

Add 1 and 3 to get 4.

$$6x+7=4+2\sqrt{3x+3}+3x$$

Subtract 4+3x from both sides of the equation.

$$6x+7-\left(4+3x\right)=2\sqrt{3x+3}$$

To find the opposite of 4+3x, find the opposite of each term.

$$6x+7-4-3x=2\sqrt{3x+3}$$

Subtract 4 from 7 to get 3.

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

Combine 6x and -3x to get 3x.

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

Square both sides of the equation.

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

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

$$9x^{2}+18x+9=\left(2\sqrt{3x+3}\right)^{2}$$

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

$$9x^{2}+18x+9=2^{2}\left(\sqrt{3x+3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}+18x+9=4\left(\sqrt{3x+3}\right)^{2}$$

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

$$9x^{2}+18x+9=4\left(3x+3\right)$$

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

$$9x^{2}+18x+9=12x+12$$

Subtract 12x from both sides.

$$9x^{2}+18x+9-12x=12$$

Combine 18x and -12x to get 6x.

$$9x^{2}+6x+9=12$$

Subtract 12 from both sides.

$$9x^{2}+6x+9-12=0$$

Subtract 12 from 9 to get -3.

$$9x^{2}+6x-3=0$$

Divide both sides by 3.

$$3x^{2}+2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-1\right)=-3$$

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.

$$a=-1 b=3$$

Rewrite 3x^{2}+2x-1 as \left(3x^{2}-x\right)+\left(3x-1\right).

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

Factor out x in 3x^{2}-x.

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

Factor out common term 3x-1 by using distributive property.

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

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

$$x=\frac{1}{3} x=-1$$

Substitute \frac{1}{3} for x in the equation \sqrt{6x+7}-\sqrt{3x+3}=1.

$$\sqrt{6\times \frac{1}{3}+7}-\sqrt{3\times \frac{1}{3}+3}=1$$

Simplify. The value x=\frac{1}{3} satisfies the equation.

$$1=1$$

Substitute -1 for x in the equation \sqrt{6x+7}-\sqrt{3x+3}=1.

$$\sqrt{6\left(-1\right)+7}-\sqrt{3\left(-1\right)+3}=1$$

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

$$1=1$$

List all solutions of \sqrt{6x+7}=\sqrt{3x+3}+1.

$$x=\frac{1}{3} x=-1$$