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

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

Solve for x

$$x=10$$

$$x=1$$

Solution Steps

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

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

Square both sides of the equation.

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

Calculate \sqrt{3x+6} to the power of 2 and get 3x+6.

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

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

$$3x+6=9+6\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

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

$$3x+6=9+6\sqrt{x-1}+x-1$$

Subtract 1 from 9 to get 8.

$$3x+6=8+6\sqrt{x-1}+x$$

Subtract 8+x from both sides of the equation.

$$3x+6-\left(8+x\right)=6\sqrt{x-1}$$

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

$$3x+6-8-x=6\sqrt{x-1}$$

Subtract 8 from 6 to get -2.

$$3x-2-x=6\sqrt{x-1}$$

Combine 3x and -x to get 2x.

$$2x-2=6\sqrt{x-1}$$

Square both sides of the equation.

$$\left(2x-2\right)^{2}=\left(6\sqrt{x-1}\right)^{2}$$

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

$$4x^{2}-8x+4=\left(6\sqrt{x-1}\right)^{2}$$

Expand \left(6\sqrt{x-1}\right)^{2}.

$$4x^{2}-8x+4=6^{2}\left(\sqrt{x-1}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$4x^{2}-8x+4=36\left(\sqrt{x-1}\right)^{2}$$

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

$$4x^{2}-8x+4=36\left(x-1\right)$$

Use the distributive property to multiply 36 by x-1.

$$4x^{2}-8x+4=36x-36$$

Subtract 36x from both sides.

$$4x^{2}-8x+4-36x=-36$$

Combine -8x and -36x to get -44x.

$$4x^{2}-44x+4=-36$$

Add 36 to both sides.

$$4x^{2}-44x+4+36=0$$

Add 4 and 36 to get 40.

$$4x^{2}-44x+40=0$$

Divide both sides by 4.

$$x^{2}-11x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite x^{2}-11x+10 as \left(x^{2}-10x\right)+\left(-x+10\right).

$$\left(x^{2}-10x\right)+\left(-x+10\right)$$

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

$$x\left(x-10\right)-\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-1\right)$$

To find equation solutions, solve x-10=0 and x-1=0.

$$x=10 x=1$$

Substitute 10 for x in the equation \sqrt{3x+6}-\sqrt{x-1}=3.

$$\sqrt{3\times 10+6}-\sqrt{10-1}=3$$

Simplify. The value x=10 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=1 satisfies the equation.

$$3=3$$

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

$$x=10 x=1$$