Solve sqrt{5-2x}-sqrt{x+6}=sqrt{x+3}

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

Solve for x

$$x=-2$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5-2x}-\sqrt{x+6}\right)^{2}.

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

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

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

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

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

Combine -2x and x to get -x.

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

Add 5 and 6 to get 11.

$$11-x-2\sqrt{5-2x}\sqrt{x+6}=\left(\sqrt{x+3}\right)^{2}$$

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

$$11-x-2\sqrt{5-2x}\sqrt{x+6}=x+3$$

Subtract 11-x from both sides of the equation.

$$-2\sqrt{5-2x}\sqrt{x+6}=x+3-\left(11-x\right)$$

To find the opposite of 11-x, find the opposite of each term.

$$-2\sqrt{5-2x}\sqrt{x+6}=x+3-11+x$$

Subtract 11 from 3 to get -8.

$$-2\sqrt{5-2x}\sqrt{x+6}=x-8+x$$

Combine x and x to get 2x.

$$-2\sqrt{5-2x}\sqrt{x+6}=2x-8$$

Square both sides of the equation.

$$\left(-2\sqrt{5-2x}\sqrt{x+6}\right)^{2}=\left(2x-8\right)^{2}$$

Expand \left(-2\sqrt{5-2x}\sqrt{x+6}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{5-2x}\right)^{2}\left(\sqrt{x+6}\right)^{2}=\left(2x-8\right)^{2}$$

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

$$4\left(\sqrt{5-2x}\right)^{2}\left(\sqrt{x+6}\right)^{2}=\left(2x-8\right)^{2}$$

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

$$4\left(5-2x\right)\left(\sqrt{x+6}\right)^{2}=\left(2x-8\right)^{2}$$

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

$$4\left(5-2x\right)\left(x+6\right)=\left(2x-8\right)^{2}$$

Use the distributive property to multiply 4 by 5-2x.

$$\left(20-8x\right)\left(x+6\right)=\left(2x-8\right)^{2}$$

Apply the distributive property by multiplying each term of 20-8x by each term of x+6.

$$20x+120-8x^{2}-48x=\left(2x-8\right)^{2}$$

Combine 20x and -48x to get -28x.

$$-28x+120-8x^{2}=\left(2x-8\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-8\right)^{2}.

$$-28x+120-8x^{2}=4x^{2}-32x+64$$

Subtract 4x^{2} from both sides.

$$-28x+120-8x^{2}-4x^{2}=-32x+64$$

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$-28x+120-12x^{2}=-32x+64$$

Add 32x to both sides.

$$-28x+120-12x^{2}+32x=64$$

Combine -28x and 32x to get 4x.

$$4x+120-12x^{2}=64$$

Subtract 64 from both sides.

$$4x+120-12x^{2}-64=0$$

Subtract 64 from 120 to get 56.

$$4x+56-12x^{2}=0$$

Divide both sides by 4.

$$x+14-3x^{2}=0$$

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

$$-3x^{2}+x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=1 ab=-3\times 14=-42$$

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. List all such integer pairs that give product -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=7 b=-6$$

Rewrite -3x^{2}+x+14 as \left(-3x^{2}+7x\right)+\left(-6x+14\right).

$$\left(-3x^{2}+7x\right)+\left(-6x+14\right)$$

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

$$-x\left(3x-7\right)-2\left(3x-7\right)$$

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(-x-2\right)$$

To find equation solutions, solve 3x-7=0 and -x-2=0.

$$x=\frac{7}{3} x=-2$$

Substitute \frac{7}{3} for x in the equation \sqrt{5-2x}-\sqrt{x+6}=\sqrt{x+3}.

$$\sqrt{5-2\times \frac{7}{3}}-\sqrt{\frac{7}{3}+6}=\sqrt{\frac{7}{3}+3}$$

Simplify. The value x=\frac{7}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{4}{3}\times 3^{\frac{1}{2}}=\frac{4}{3}\times 3^{\frac{1}{2}}$$

Substitute -2 for x in the equation \sqrt{5-2x}-\sqrt{x+6}=\sqrt{x+3}.

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

Simplify. The value x=-2 satisfies the equation.

$$1=1$$

Equation -\sqrt{x+6}+\sqrt{5-2x}=\sqrt{x+3} has a unique solution.

$$x=-2$$