Solve sqrt{4+3x}+sqrt{x+2}=2

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

Solve for x

$$x=-1$$

Solution Steps

Subtract \sqrt{x+2} from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{4+3x} to the power of 2 and get 4+3x.

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

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

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

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

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

Add 4 and 2 to get 6.

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

Subtract 6+x from both sides of the equation.

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

To find the opposite of 6+x, find the opposite of each term.

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

Subtract 6 from 4 to get -2.

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

Combine 3x and -x to get 2x.

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

Square both sides of the equation.

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

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

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

Expand \left(-4\sqrt{x+2}\right)^{2}.

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

Calculate -4 to the power of 2 and get 16.

$$4-8x+4x^{2}=16\left(\sqrt{x+2}\right)^{2}$$

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

$$4-8x+4x^{2}=16\left(x+2\right)$$

Use the distributive property to multiply 16 by x+2.

$$4-8x+4x^{2}=16x+32$$

Subtract 16x from both sides.

$$4-8x+4x^{2}-16x=32$$

Combine -8x and -16x to get -24x.

$$4-24x+4x^{2}=32$$

Subtract 32 from both sides.

$$4-24x+4x^{2}-32=0$$

Subtract 32 from 4 to get -28.

$$-28-24x+4x^{2}=0$$

Divide both sides by 4.

$$-7-6x+x^{2}=0$$

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

$$x^{2}-6x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-7\right)=-7$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-7 b=1$$

Rewrite x^{2}-6x-7 as \left(x^{2}-7x\right)+\left(x-7\right).

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

Factor out x in x^{2}-7x.

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

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

$$\left(x-7\right)\left(x+1\right)$$

To find equation solutions, solve x-7=0 and x+1=0.

$$x=7 x=-1$$

Substitute 7 for x in the equation \sqrt{4+3x}+\sqrt{x+2}=2.

$$\sqrt{4+3\times 7}+\sqrt{7+2}=2$$

Simplify. The value x=7 does not satisfy the equation.

$$8=2$$

Substitute -1 for x in the equation \sqrt{4+3x}+\sqrt{x+2}=2.

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

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

$$2=2$$

Equation \sqrt{3x+4}=-\sqrt{x+2}+2 has a unique solution.

$$x=-1$$