Solve sqrt{2x+7}+sqrt{3x-18}=sqrt{7x+1}

$$\sqrt{ 2x+7 } + \sqrt{ 3x-18 } = \sqrt{ 7x+1 }$$

Solve for x

$$x=9$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+7}+\sqrt{3x-18}\right)^{2}=\left(\sqrt{7x+1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2x+7}+\sqrt{3x-18}\right)^{2}.

$$\left(\sqrt{2x+7}\right)^{2}+2\sqrt{2x+7}\sqrt{3x-18}+\left(\sqrt{3x-18}\right)^{2}=\left(\sqrt{7x+1}\right)^{2}$$

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

$$2x+7+2\sqrt{2x+7}\sqrt{3x-18}+\left(\sqrt{3x-18}\right)^{2}=\left(\sqrt{7x+1}\right)^{2}$$

Calculate \sqrt{3x-18} to the power of 2 and get 3x-18.

$$2x+7+2\sqrt{2x+7}\sqrt{3x-18}+3x-18=\left(\sqrt{7x+1}\right)^{2}$$

Combine 2x and 3x to get 5x.

$$5x+7+2\sqrt{2x+7}\sqrt{3x-18}-18=\left(\sqrt{7x+1}\right)^{2}$$

Subtract 18 from 7 to get -11.

$$5x-11+2\sqrt{2x+7}\sqrt{3x-18}=\left(\sqrt{7x+1}\right)^{2}$$

Calculate \sqrt{7x+1} to the power of 2 and get 7x+1.

$$5x-11+2\sqrt{2x+7}\sqrt{3x-18}=7x+1$$

Subtract 5x-11 from both sides of the equation.

$$2\sqrt{2x+7}\sqrt{3x-18}=7x+1-\left(5x-11\right)$$

To find the opposite of 5x-11, find the opposite of each term.

$$2\sqrt{2x+7}\sqrt{3x-18}=7x+1-5x+11$$

Combine 7x and -5x to get 2x.

$$2\sqrt{2x+7}\sqrt{3x-18}=2x+1+11$$

Add 1 and 11 to get 12.

$$2\sqrt{2x+7}\sqrt{3x-18}=2x+12$$

Square both sides of the equation.

$$\left(2\sqrt{2x+7}\sqrt{3x-18}\right)^{2}=\left(2x+12\right)^{2}$$

Expand \left(2\sqrt{2x+7}\sqrt{3x-18}\right)^{2}.

$$2^{2}\left(\sqrt{2x+7}\right)^{2}\left(\sqrt{3x-18}\right)^{2}=\left(2x+12\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{2x+7}\right)^{2}\left(\sqrt{3x-18}\right)^{2}=\left(2x+12\right)^{2}$$

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

$$4\left(2x+7\right)\left(\sqrt{3x-18}\right)^{2}=\left(2x+12\right)^{2}$$

Calculate \sqrt{3x-18} to the power of 2 and get 3x-18.

$$4\left(2x+7\right)\left(3x-18\right)=\left(2x+12\right)^{2}$$

Use the distributive property to multiply 4 by 2x+7.

$$\left(8x+28\right)\left(3x-18\right)=\left(2x+12\right)^{2}$$

Apply the distributive property by multiplying each term of 8x+28 by each term of 3x-18.

$$24x^{2}-144x+84x-504=\left(2x+12\right)^{2}$$

Combine -144x and 84x to get -60x.

$$24x^{2}-60x-504=\left(2x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+12\right)^{2}.

$$24x^{2}-60x-504=4x^{2}+48x+144$$

Subtract 4x^{2} from both sides.

$$24x^{2}-60x-504-4x^{2}=48x+144$$

Combine 24x^{2} and -4x^{2} to get 20x^{2}.

$$20x^{2}-60x-504=48x+144$$

Subtract 48x from both sides.

$$20x^{2}-60x-504-48x=144$$

Combine -60x and -48x to get -108x.

$$20x^{2}-108x-504=144$$

Subtract 144 from both sides.

$$20x^{2}-108x-504-144=0$$

Subtract 144 from -504 to get -648.

$$20x^{2}-108x-648=0$$

Divide both sides by 4.

$$5x^{2}-27x-162=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-162. To find a and b, set up a system to be solved.

$$a+b=-27 ab=5\left(-162\right)=-810$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -810.

$$1,-810 2,-405 3,-270 5,-162 6,-135 9,-90 10,-81 15,-54 18,-45 27,-30$$

Calculate the sum for each pair.

$$1-810=-809 2-405=-403 3-270=-267 5-162=-157 6-135=-129 9-90=-81 10-81=-71 15-54=-39 18-45=-27 27-30=-3$$

The solution is the pair that gives sum -27.

$$a=-45 b=18$$

Rewrite 5x^{2}-27x-162 as \left(5x^{2}-45x\right)+\left(18x-162\right).

$$\left(5x^{2}-45x\right)+\left(18x-162\right)$$

Factor out 5x in the first and 18 in the second group.

$$5x\left(x-9\right)+18\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(5x+18\right)$$

To find equation solutions, solve x-9=0 and 5x+18=0.

$$x=9 x=-\frac{18}{5}$$

Substitute -\frac{18}{5} for x in the equation \sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}. The expression \sqrt{2\left(-\frac{18}{5}\right)+7} is undefined because the radicand cannot be negative.

$$\sqrt{2\left(-\frac{18}{5}\right)+7}+\sqrt{3\left(-\frac{18}{5}\right)-18}=\sqrt{7\left(-\frac{18}{5}\right)+1}$$

Substitute 9 for x in the equation \sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}.

$$\sqrt{2\times 9+7}+\sqrt{3\times 9-18}=\sqrt{7\times 9+1}$$

Simplify. The value x=9 satisfies the equation.

$$8=8$$

Equation \sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1} has a unique solution.

$$x=9$$