Solve 30-(x+1)-(16-x)=sqrt{(x+1)^2+(16-x)^2}

$$30 - ( x + 1 ) - ( 16 - x ) = \sqrt { ( x + 1 ) ^ { 2 } + ( 16 - x ) ^ { 2 } }$$

Solve for x

$$x=11$$

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(30-\left(x+1\right)-\left(16-x\right)\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

To find the opposite of x+1, find the opposite of each term.

$$\left(30-x-1-\left(16-x\right)\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

Subtract 1 from 30 to get 29.

$$\left(29-x-\left(16-x\right)\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

To find the opposite of 16-x, find the opposite of each term.

$$\left(29-x-16+x\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

Subtract 16 from 29 to get 13.

$$\left(13-x+x\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

Combine -x and x to get 0.

$$13^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

Calculate 13 to the power of 2 and get 169.

$$169=\left(\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}\right)^{2}$$

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

$$169=\left(\sqrt{x^{2}+2x+1+\left(16-x\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16-x\right)^{2}.

$$169=\left(\sqrt{x^{2}+2x+1+256-32x+x^{2}}\right)^{2}$$

Add 1 and 256 to get 257.

$$169=\left(\sqrt{x^{2}+2x+257-32x+x^{2}}\right)^{2}$$

Combine 2x and -32x to get -30x.

$$169=\left(\sqrt{x^{2}-30x+257+x^{2}}\right)^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$169=\left(\sqrt{2x^{2}-30x+257}\right)^{2}$$

Calculate \sqrt{2x^{2}-30x+257} to the power of 2 and get 2x^{2}-30x+257.

$$169=2x^{2}-30x+257$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-30x+257=169$$

Subtract 169 from both sides.

$$2x^{2}-30x+257-169=0$$

Subtract 169 from 257 to get 88.

$$2x^{2}-30x+88=0$$

Divide both sides by 2.

$$x^{2}-15x+44=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+44. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 44=44$$

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 44.

$$-1,-44 -2,-22 -4,-11$$

Calculate the sum for each pair.

$$-1-44=-45 -2-22=-24 -4-11=-15$$

The solution is the pair that gives sum -15.

$$a=-11 b=-4$$

Rewrite x^{2}-15x+44 as \left(x^{2}-11x\right)+\left(-4x+44\right).

$$\left(x^{2}-11x\right)+\left(-4x+44\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-11\right)-4\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x-4\right)$$

To find equation solutions, solve x-11=0 and x-4=0.

$$x=11 x=4$$

Substitute 11 for x in the equation 30-\left(x+1\right)-\left(16-x\right)=\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}.

$$30-\left(11+1\right)-\left(16-11\right)=\sqrt{\left(11+1\right)^{2}+\left(16-11\right)^{2}}$$

Simplify. The value x=11 satisfies the equation.

$$13=13$$

Substitute 4 for x in the equation 30-\left(x+1\right)-\left(16-x\right)=\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}.

$$30-\left(4+1\right)-\left(16-4\right)=\sqrt{\left(4+1\right)^{2}+\left(16-4\right)^{2}}$$

Simplify. The value x=4 satisfies the equation.

$$13=13$$

List all solutions of -\left(x+1\right)-\left(16-x\right)+30=\sqrt{\left(x+1\right)^{2}+\left(16-x\right)^{2}}.

$$x=11 x=4$$