Solve sqrt{(x+2)^2+9}+sqrt{(x-1)^2+1}=5

$$\sqrt { ( x + 2 ) ^ { 2 } + 9 } + \sqrt { ( x - 1 ) ^ { 2 } + 1 } = 5$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Solution Steps

Subtract \sqrt{\left(x-1\right)^{2}+1} from both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}+9}=5-\sqrt{\left(x-1\right)^{2}+1}$$

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

$$\sqrt{x^{2}+4x+4+9}=5-\sqrt{\left(x-1\right)^{2}+1}$$

Add 4 and 9 to get 13.

$$\sqrt{x^{2}+4x+13}=5-\sqrt{\left(x-1\right)^{2}+1}$$

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

$$\sqrt{x^{2}+4x+13}=5-\sqrt{x^{2}-2x+1+1}$$

Add 1 and 1 to get 2.

$$\sqrt{x^{2}+4x+13}=5-\sqrt{x^{2}-2x+2}$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+4x+13}\right)^{2}=\left(5-\sqrt{x^{2}-2x+2}\right)^{2}$$

Calculate \sqrt{x^{2}+4x+13} to the power of 2 and get x^{2}+4x+13.

$$x^{2}+4x+13=\left(5-\sqrt{x^{2}-2x+2}\right)^{2}$$

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

$$x^{2}+4x+13=25-10\sqrt{x^{2}-2x+2}+\left(\sqrt{x^{2}-2x+2}\right)^{2}$$

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

$$x^{2}+4x+13=25-10\sqrt{x^{2}-2x+2}+x^{2}-2x+2$$

Add 25 and 2 to get 27.

$$x^{2}+4x+13=27-10\sqrt{x^{2}-2x+2}+x^{2}-2x$$

Subtract 27+x^{2}-2x from both sides of the equation.

$$x^{2}+4x+13-\left(27+x^{2}-2x\right)=-10\sqrt{x^{2}-2x+2}$$

To find the opposite of 27+x^{2}-2x, find the opposite of each term.

$$x^{2}+4x+13-27-x^{2}+2x=-10\sqrt{x^{2}-2x+2}$$

Subtract 27 from 13 to get -14.

$$x^{2}+4x-14-x^{2}+2x=-10\sqrt{x^{2}-2x+2}$$

Combine x^{2} and -x^{2} to get 0.

$$4x-14+2x=-10\sqrt{x^{2}-2x+2}$$

Combine 4x and 2x to get 6x.

$$6x-14=-10\sqrt{x^{2}-2x+2}$$

Square both sides of the equation.

$$\left(6x-14\right)^{2}=\left(-10\sqrt{x^{2}-2x+2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-14\right)^{2}.

$$36x^{2}-168x+196=\left(-10\sqrt{x^{2}-2x+2}\right)^{2}$$

Expand \left(-10\sqrt{x^{2}-2x+2}\right)^{2}.

$$36x^{2}-168x+196=\left(-10\right)^{2}\left(\sqrt{x^{2}-2x+2}\right)^{2}$$

Calculate -10 to the power of 2 and get 100.

$$36x^{2}-168x+196=100\left(\sqrt{x^{2}-2x+2}\right)^{2}$$

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

$$36x^{2}-168x+196=100\left(x^{2}-2x+2\right)$$

Use the distributive property to multiply 100 by x^{2}-2x+2.

$$36x^{2}-168x+196=100x^{2}-200x+200$$

Subtract 100x^{2} from both sides.

$$36x^{2}-168x+196-100x^{2}=-200x+200$$

Combine 36x^{2} and -100x^{2} to get -64x^{2}.

$$-64x^{2}-168x+196=-200x+200$$

Add 200x to both sides.

$$-64x^{2}-168x+196+200x=200$$

Combine -168x and 200x to get 32x.

$$-64x^{2}+32x+196=200$$

Subtract 200 from both sides.

$$-64x^{2}+32x+196-200=0$$

Subtract 200 from 196 to get -4.

$$-64x^{2}+32x-4=0$$

Divide both sides by 4.

$$-16x^{2}+8x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=8 ab=-16\left(-1\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite -16x^{2}+8x-1 as \left(-16x^{2}+4x\right)+\left(4x-1\right).

$$\left(-16x^{2}+4x\right)+\left(4x-1\right)$$

Factor out -4x in -16x^{2}+4x.

$$-4x\left(4x-1\right)+4x-1$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(-4x+1\right)$$

To find equation solutions, solve 4x-1=0 and -4x+1=0.

$$x=\frac{1}{4} x=\frac{1}{4}$$

Substitute \frac{1}{4} for x in the equation \sqrt{\left(x+2\right)^{2}+9}+\sqrt{\left(x-1\right)^{2}+1}=5.

$$\sqrt{\left(\frac{1}{4}+2\right)^{2}+9}+\sqrt{\left(\frac{1}{4}-1\right)^{2}+1}=5$$

Simplify. The value x=\frac{1}{4} satisfies the equation.

$$5=5$$

Substitute \frac{1}{4} for x in the equation \sqrt{\left(x+2\right)^{2}+9}+\sqrt{\left(x-1\right)^{2}+1}=5.

$$\sqrt{\left(\frac{1}{4}+2\right)^{2}+9}+\sqrt{\left(\frac{1}{4}-1\right)^{2}+1}=5$$

Simplify. The value x=\frac{1}{4} satisfies the equation.

$$5=5$$

List all solutions of \sqrt{x^{2}+4x+13}=-\sqrt{x^{2}-2x+2}+5.

$$x=\frac{1}{4} x=\frac{1}{4}$$