Solve (sqrt{x+6}=sqrt{11-x}-3)

$$( \sqrt { x + 6 } = \sqrt { 11 - x } - 3 )$$

Solve for x

$$x=-5$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{11-x}-3\right)^{2}.

$$x+6=\left(\sqrt{11-x}\right)^{2}-6\sqrt{11-x}+9$$

Calculate \sqrt{11-x} to the power of 2 and get 11-x.

$$x+6=11-x-6\sqrt{11-x}+9$$

Add 11 and 9 to get 20.

$$x+6=20-x-6\sqrt{11-x}$$

Subtract 20-x from both sides of the equation.

$$x+6-\left(20-x\right)=-6\sqrt{11-x}$$

To find the opposite of 20-x, find the opposite of each term.

$$x+6-20+x=-6\sqrt{11-x}$$

Subtract 20 from 6 to get -14.

$$x-14+x=-6\sqrt{11-x}$$

Combine x and x to get 2x.

$$2x-14=-6\sqrt{11-x}$$

Square both sides of the equation.

$$\left(2x-14\right)^{2}=\left(-6\sqrt{11-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-14\right)^{2}.

$$4x^{2}-56x+196=\left(-6\sqrt{11-x}\right)^{2}$$

Expand \left(-6\sqrt{11-x}\right)^{2}.

$$4x^{2}-56x+196=\left(-6\right)^{2}\left(\sqrt{11-x}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$4x^{2}-56x+196=36\left(\sqrt{11-x}\right)^{2}$$

Calculate \sqrt{11-x} to the power of 2 and get 11-x.

$$4x^{2}-56x+196=36\left(11-x\right)$$

Use the distributive property to multiply 36 by 11-x.

$$4x^{2}-56x+196=396-36x$$

Subtract 396 from both sides.

$$4x^{2}-56x+196-396=-36x$$

Subtract 396 from 196 to get -200.

$$4x^{2}-56x-200=-36x$$

Add 36x to both sides.

$$4x^{2}-56x-200+36x=0$$

Combine -56x and 36x to get -20x.

$$4x^{2}-20x-200=0$$

Divide both sides by 4.

$$x^{2}-5x-50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-50. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-50\right)=-50$$

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 -50.

$$1,-50 2,-25 5,-10$$

Calculate the sum for each pair.

$$1-50=-49 2-25=-23 5-10=-5$$

The solution is the pair that gives sum -5.

$$a=-10 b=5$$

Rewrite x^{2}-5x-50 as \left(x^{2}-10x\right)+\left(5x-50\right).

$$\left(x^{2}-10x\right)+\left(5x-50\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-10\right)+5\left(x-10\right)$$

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

$$\left(x-10\right)\left(x+5\right)$$

To find equation solutions, solve x-10=0 and x+5=0.

$$x=10 x=-5$$

Substitute 10 for x in the equation \sqrt{x+6}=\sqrt{11-x}-3.

$$\sqrt{10+6}=\sqrt{11-10}-3$$

Simplify. The value x=10 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-2$$

Substitute -5 for x in the equation \sqrt{x+6}=\sqrt{11-x}-3.

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

Simplify. The value x=-5 satisfies the equation.

$$1=1$$

Equation \sqrt{x+6}=\sqrt{11-x}-3 has a unique solution.

$$x=-5$$