Solve sqrt{7x+46}-sqrt{x+31}=3

$$\sqrt { 7 x + 46 } - \sqrt { x + 31 } = 3$$

Solve for x

$$x=5$$

Solution Steps

Subtract -\sqrt{x+31} from both sides of the equation.

$$\sqrt{7x+46}=3+\sqrt{x+31}$$

Square both sides of the equation.

$$\left(\sqrt{7x+46}\right)^{2}=\left(3+\sqrt{x+31}\right)^{2}$$

Calculate \sqrt{7x+46} to the power of 2 and get 7x+46.

$$7x+46=\left(3+\sqrt{x+31}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{x+31}\right)^{2}.

$$7x+46=9+6\sqrt{x+31}+\left(\sqrt{x+31}\right)^{2}$$

Calculate \sqrt{x+31} to the power of 2 and get x+31.

$$7x+46=9+6\sqrt{x+31}+x+31$$

Add 9 and 31 to get 40.

$$7x+46=40+6\sqrt{x+31}+x$$

Subtract 40+x from both sides of the equation.

$$7x+46-\left(40+x\right)=6\sqrt{x+31}$$

To find the opposite of 40+x, find the opposite of each term.

$$7x+46-40-x=6\sqrt{x+31}$$

Subtract 40 from 46 to get 6.

$$7x+6-x=6\sqrt{x+31}$$

Combine 7x and -x to get 6x.

$$6x+6=6\sqrt{x+31}$$

Square both sides of the equation.

$$\left(6x+6\right)^{2}=\left(6\sqrt{x+31}\right)^{2}$$

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

$$36x^{2}+72x+36=\left(6\sqrt{x+31}\right)^{2}$$

Expand \left(6\sqrt{x+31}\right)^{2}.

$$36x^{2}+72x+36=6^{2}\left(\sqrt{x+31}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36x^{2}+72x+36=36\left(\sqrt{x+31}\right)^{2}$$

Calculate \sqrt{x+31} to the power of 2 and get x+31.

$$36x^{2}+72x+36=36\left(x+31\right)$$

Use the distributive property to multiply 36 by x+31.

$$36x^{2}+72x+36=36x+1116$$

Subtract 36x from both sides.

$$36x^{2}+72x+36-36x=1116$$

Combine 72x and -36x to get 36x.

$$36x^{2}+36x+36=1116$$

Subtract 1116 from both sides.

$$36x^{2}+36x+36-1116=0$$

Subtract 1116 from 36 to get -1080.

$$36x^{2}+36x-1080=0$$

Divide both sides by 36.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite x^{2}+x-30 as \left(x^{2}-5x\right)+\left(6x-30\right).

$$\left(x^{2}-5x\right)+\left(6x-30\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-5\right)+6\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+6\right)$$

To find equation solutions, solve x-5=0 and x+6=0.

$$x=5 x=-6$$

Substitute 5 for x in the equation \sqrt{7x+46}-\sqrt{x+31}=3.

$$\sqrt{7\times 5+46}-\sqrt{5+31}=3$$

Simplify. The value x=5 satisfies the equation.

$$3=3$$

Substitute -6 for x in the equation \sqrt{7x+46}-\sqrt{x+31}=3.

$$\sqrt{7\left(-6\right)+46}-\sqrt{-6+31}=3$$

Simplify. The value x=-6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

Substitute 5 for x in the equation \sqrt{7x+46}-\sqrt{x+31}=3.

$$\sqrt{7\times 5+46}-\sqrt{5+31}=3$$

Simplify. The value x=5 satisfies the equation.

$$3=3$$

Equation \sqrt{7x+46}=\sqrt{x+31}+3 has a unique solution.

$$x=5$$