Solve sqrt{10-3x}=2+sqrt{x+6}

$$\sqrt { 10 - 3 x } = 2 + \sqrt { x + 6 }$$

Solve for x

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{10-3x}\right)^{2}=\left(2+\sqrt{x+6}\right)^{2}$$

Calculate \sqrt{10-3x} to the power of 2 and get 10-3x.

$$10-3x=\left(2+\sqrt{x+6}\right)^{2}$$

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

$$10-3x=4+4\sqrt{x+6}+\left(\sqrt{x+6}\right)^{2}$$

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

$$10-3x=4+4\sqrt{x+6}+x+6$$

Add 4 and 6 to get 10.

$$10-3x=10+4\sqrt{x+6}+x$$

Subtract 10+x from both sides of the equation.

$$10-3x-\left(10+x\right)=4\sqrt{x+6}$$

To find the opposite of 10+x, find the opposite of each term.

$$10-3x-10-x=4\sqrt{x+6}$$

Subtract 10 from 10 to get 0.

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

Combine -3x and -x to get -4x.

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

Square both sides of the equation.

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

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

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

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

$$16x^{2}=\left(4\sqrt{x+6}\right)^{2}$$

Expand \left(4\sqrt{x+6}\right)^{2}.

$$16x^{2}=4^{2}\left(\sqrt{x+6}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}=16\left(\sqrt{x+6}\right)^{2}$$

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

$$16x^{2}=16\left(x+6\right)$$

Use the distributive property to multiply 16 by x+6.

$$16x^{2}=16x+96$$

Subtract 16x from both sides.

$$16x^{2}-16x=96$$

Subtract 96 from both sides.

$$16x^{2}-16x-96=0$$

Divide both sides by 16.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

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

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. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

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

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

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

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

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

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

$$x=3 x=-2$$

Substitute 3 for x in the equation \sqrt{10-3x}=2+\sqrt{x+6}.

$$\sqrt{10-3\times 3}=2+\sqrt{3+6}$$

Simplify. The value x=3 does not satisfy the equation.

$$1=5$$

Substitute -2 for x in the equation \sqrt{10-3x}=2+\sqrt{x+6}.

$$\sqrt{10-3\left(-2\right)}=2+\sqrt{-2+6}$$

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

$$4=4$$

Equation \sqrt{10-3x}=\sqrt{x+6}+2 has a unique solution.

$$x=-2$$