Solve (sqrt{x+2}+sqrt{18-x})^2=(sqrt{10+13x})^2

$$( \sqrt { x + 2 } + \sqrt { 18 - x } ) ^ { 2 } = ( \sqrt { 10 + 13 x } ) ^ { 2 }$$

Solve for x

$$x=2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+2}+\sqrt{18-x}\right)^{2}.

$$\left(\sqrt{x+2}\right)^{2}+2\sqrt{x+2}\sqrt{18-x}+\left(\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2}$$

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

$$x+2+2\sqrt{x+2}\sqrt{18-x}+\left(\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2}$$

Calculate \sqrt{18-x} to the power of 2 and get 18-x.

$$x+2+2\sqrt{x+2}\sqrt{18-x}+18-x=\left(\sqrt{10+13x}\right)^{2}$$

Add 2 and 18 to get 20.

$$x+20+2\sqrt{x+2}\sqrt{18-x}-x=\left(\sqrt{10+13x}\right)^{2}$$

Combine x and -x to get 0.

$$20+2\sqrt{x+2}\sqrt{18-x}=\left(\sqrt{10+13x}\right)^{2}$$

Calculate \sqrt{10+13x} to the power of 2 and get 10+13x.

$$20+2\sqrt{x+2}\sqrt{18-x}=10+13x$$

Subtract 13x from both sides.

$$20+2\sqrt{x+2}\sqrt{18-x}-13x=10$$

Subtract 20 from both sides.

$$2\sqrt{x+2}\sqrt{18-x}-13x=10-20$$

Subtract 20 from 10 to get -10.

$$2\sqrt{x+2}\sqrt{18-x}-13x=-10$$

Subtract -13x from both sides of the equation.

$$2\sqrt{x+2}\sqrt{18-x}=-10+13x$$

Square both sides of the equation.

$$\left(2\sqrt{x+2}\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

Expand \left(2\sqrt{x+2}\sqrt{18-x}\right)^{2}.

$$2^{2}\left(\sqrt{x+2}\right)^{2}\left(\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+2}\right)^{2}\left(\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

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

$$4\left(x+2\right)\left(\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

Calculate \sqrt{18-x} to the power of 2 and get 18-x.

$$4\left(x+2\right)\left(18-x\right)=\left(13x-10\right)^{2}$$

Use the distributive property to multiply 4 by x+2.

$$\left(4x+8\right)\left(18-x\right)=\left(13x-10\right)^{2}$$

Use the distributive property to multiply 4x+8 by 18-x and combine like terms.

$$64x-4x^{2}+144=\left(13x-10\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(13x-10\right)^{2}.

$$64x-4x^{2}+144=169x^{2}-260x+100$$

Subtract 169x^{2} from both sides.

$$64x-4x^{2}+144-169x^{2}=-260x+100$$

Combine -4x^{2} and -169x^{2} to get -173x^{2}.

$$64x-173x^{2}+144=-260x+100$$

Add 260x to both sides.

$$64x-173x^{2}+144+260x=100$$

Combine 64x and 260x to get 324x.

$$324x-173x^{2}+144=100$$

Subtract 100 from both sides.

$$324x-173x^{2}+144-100=0$$

Subtract 100 from 144 to get 44.

$$324x-173x^{2}+44=0$$

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

$$-173x^{2}+324x+44=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -173x^{2}+ax+bx+44. To find a and b, set up a system to be solved.

$$a+b=324 ab=-173\times 44=-7612$$

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 -7612.

$$-1,7612 -2,3806 -4,1903 -11,692 -22,346 -44,173$$

Calculate the sum for each pair.

$$-1+7612=7611 -2+3806=3804 -4+1903=1899 -11+692=681 -22+346=324 -44+173=129$$

The solution is the pair that gives sum 324.

$$a=346 b=-22$$

Rewrite -173x^{2}+324x+44 as \left(-173x^{2}+346x\right)+\left(-22x+44\right).

$$\left(-173x^{2}+346x\right)+\left(-22x+44\right)$$

Factor out 173x in the first and 22 in the second group.

$$173x\left(-x+2\right)+22\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(173x+22\right)$$

To find equation solutions, solve -x+2=0 and 173x+22=0.

$$x=2 x=-\frac{22}{173}$$

Substitute 2 for x in the equation \left(\sqrt{x+2}+\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2}.

$$\left(\sqrt{2+2}+\sqrt{18-2}\right)^{2}=\left(\sqrt{10+13\times 2}\right)^{2}$$

Simplify. The value x=2 satisfies the equation.

$$36=36$$

Substitute -\frac{22}{173} for x in the equation \left(\sqrt{x+2}+\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2}.

$$\left(\sqrt{-\frac{22}{173}+2}+\sqrt{18-\left(-\frac{22}{173}\right)}\right)^{2}=\left(\sqrt{10+13\left(-\frac{22}{173}\right)}\right)^{2}$$

Simplify. The value x=-\frac{22}{173} does not satisfy the equation.

$$\frac{5476}{173}=\frac{1444}{173}$$

Equation 2\sqrt{18-x}\sqrt{x+2}=13x-10 has a unique solution.

$$x=2$$