Solve sqrt{x+2}-sqrt{3x-5}=-1

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

Solve for x

$$x=7$$

Solution Steps

Subtract -\sqrt{3x-5} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-1+\sqrt{3x-5}\right)^{2}.

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

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

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

Subtract 5 from 1 to get -4.

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

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

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

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

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

Add 2 and 4 to get 6.

$$x+6-3x=-2\sqrt{3x-5}$$

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

$$-2x+6=-2\sqrt{3x-5}$$

Square both sides of the equation.

$$\left(-2x+6\right)^{2}=\left(-2\sqrt{3x-5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x+6\right)^{2}.

$$4x^{2}-24x+36=\left(-2\sqrt{3x-5}\right)^{2}$$

Expand \left(-2\sqrt{3x-5}\right)^{2}.

$$4x^{2}-24x+36=\left(-2\right)^{2}\left(\sqrt{3x-5}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4x^{2}-24x+36=4\left(\sqrt{3x-5}\right)^{2}$$

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

$$4x^{2}-24x+36=4\left(3x-5\right)$$

Use the distributive property to multiply 4 by 3x-5.

$$4x^{2}-24x+36=12x-20$$

Subtract 12x from both sides.

$$4x^{2}-24x+36-12x=-20$$

Combine -24x and -12x to get -36x.

$$4x^{2}-36x+36=-20$$

Add 20 to both sides.

$$4x^{2}-36x+36+20=0$$

Add 36 and 20 to get 56.

$$4x^{2}-36x+56=0$$

Divide both sides by 4.

$$x^{2}-9x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 14=14$$

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 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

Rewrite x^{2}-9x+14 as \left(x^{2}-7x\right)+\left(-2x+14\right).

$$\left(x^{2}-7x\right)+\left(-2x+14\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-7\right)-2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x-2\right)$$

To find equation solutions, solve x-7=0 and x-2=0.

$$x=7 x=2$$

Substitute 7 for x in the equation \sqrt{x+2}-\sqrt{3x-5}=-1.

$$\sqrt{7+2}-\sqrt{3\times 7-5}=-1$$

Simplify. The value x=7 satisfies the equation.

$$-1=-1$$

Substitute 2 for x in the equation \sqrt{x+2}-\sqrt{3x-5}=-1.

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

Simplify. The value x=2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Substitute 7 for x in the equation \sqrt{x+2}-\sqrt{3x-5}=-1.

$$\sqrt{7+2}-\sqrt{3\times 7-5}=-1$$

Simplify. The value x=7 satisfies the equation.

$$-1=-1$$

Equation \sqrt{x+2}=\sqrt{3x-5}-1 has a unique solution.

$$x=7$$