Solve sqrt{x+5}+sqrt{20-x}=7

$$\sqrt { x + 5 } + \sqrt { 20 - x } = 7$$

Solve for x

$$x=11$$

$$x=4$$

Solution Steps

Subtract \sqrt{20-x} from both sides of the equation.

$$\sqrt{x+5}=7-\sqrt{20-x}$$

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(7-\sqrt{20-x}\right)^{2}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$x+5=\left(7-\sqrt{20-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-\sqrt{20-x}\right)^{2}.

$$x+5=49-14\sqrt{20-x}+\left(\sqrt{20-x}\right)^{2}$$

Calculate \sqrt{20-x} to the power of 2 and get 20-x.

$$x+5=49-14\sqrt{20-x}+20-x$$

Add 49 and 20 to get 69.

$$x+5=69-14\sqrt{20-x}-x$$

Subtract 69-x from both sides of the equation.

$$x+5-\left(69-x\right)=-14\sqrt{20-x}$$

To find the opposite of 69-x, find the opposite of each term.

$$x+5-69+x=-14\sqrt{20-x}$$

Subtract 69 from 5 to get -64.

$$x-64+x=-14\sqrt{20-x}$$

Combine x and x to get 2x.

$$2x-64=-14\sqrt{20-x}$$

Square both sides of the equation.

$$\left(2x-64\right)^{2}=\left(-14\sqrt{20-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-64\right)^{2}.

$$4x^{2}-256x+4096=\left(-14\sqrt{20-x}\right)^{2}$$

Expand \left(-14\sqrt{20-x}\right)^{2}.

$$4x^{2}-256x+4096=\left(-14\right)^{2}\left(\sqrt{20-x}\right)^{2}$$

Calculate -14 to the power of 2 and get 196.

$$4x^{2}-256x+4096=196\left(\sqrt{20-x}\right)^{2}$$

Calculate \sqrt{20-x} to the power of 2 and get 20-x.

$$4x^{2}-256x+4096=196\left(20-x\right)$$

Use the distributive property to multiply 196 by 20-x.

$$4x^{2}-256x+4096=3920-196x$$

Subtract 3920 from both sides.

$$4x^{2}-256x+4096-3920=-196x$$

Subtract 3920 from 4096 to get 176.

$$4x^{2}-256x+176=-196x$$

Add 196x to both sides.

$$4x^{2}-256x+176+196x=0$$

Combine -256x and 196x to get -60x.

$$4x^{2}-60x+176=0$$

Divide both sides by 4.

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

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 \sqrt{x+5}+\sqrt{20-x}=7.

$$\sqrt{11+5}+\sqrt{20-11}=7$$

Simplify. The value x=11 satisfies the equation.

$$7=7$$

Substitute 4 for x in the equation \sqrt{x+5}+\sqrt{20-x}=7.

$$\sqrt{4+5}+\sqrt{20-4}=7$$

Simplify. The value x=4 satisfies the equation.

$$7=7$$

List all solutions of \sqrt{x+5}=-\sqrt{20-x}+7.

$$x=11 x=4$$