Solve sqrt{5x+1}-sqrt{3x-5}=sqrt{x+1}

$$\sqrt { 5 x + 1 } - \sqrt { 3 x - 5 } = \sqrt { x + 1 }$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5x+1}-\sqrt{3x-5}\right)^{2}.

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

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

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

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

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

Combine 5x and 3x to get 8x.

$$8x+1-2\sqrt{5x+1}\sqrt{3x-5}-5=\left(\sqrt{x+1}\right)^{2}$$

Subtract 5 from 1 to get -4.

$$8x-4-2\sqrt{5x+1}\sqrt{3x-5}=\left(\sqrt{x+1}\right)^{2}$$

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

$$8x-4-2\sqrt{5x+1}\sqrt{3x-5}=x+1$$

Subtract 8x-4 from both sides of the equation.

$$-2\sqrt{5x+1}\sqrt{3x-5}=x+1-\left(8x-4\right)$$

To find the opposite of 8x-4, find the opposite of each term.

$$-2\sqrt{5x+1}\sqrt{3x-5}=x+1-8x+4$$

Combine x and -8x to get -7x.

$$-2\sqrt{5x+1}\sqrt{3x-5}=-7x+1+4$$

Add 1 and 4 to get 5.

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

Square both sides of the equation.

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

Expand \left(-2\sqrt{5x+1}\sqrt{3x-5}\right)^{2}.

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

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{5x+1}\right)^{2}\left(\sqrt{3x-5}\right)^{2}=\left(-7x+5\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$4\left(5x+1\right)\left(\sqrt{3x-5}\right)^{2}=\left(-7x+5\right)^{2}$$

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

$$4\left(5x+1\right)\left(3x-5\right)=\left(-7x+5\right)^{2}$$

Use the distributive property to multiply 4 by 5x+1.

$$\left(20x+4\right)\left(3x-5\right)=\left(-7x+5\right)^{2}$$

Apply the distributive property by multiplying each term of 20x+4 by each term of 3x-5.

$$60x^{2}-100x+12x-20=\left(-7x+5\right)^{2}$$

Combine -100x and 12x to get -88x.

$$60x^{2}-88x-20=\left(-7x+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-7x+5\right)^{2}.

$$60x^{2}-88x-20=49x^{2}-70x+25$$

Subtract 49x^{2} from both sides.

$$60x^{2}-88x-20-49x^{2}=-70x+25$$

Combine 60x^{2} and -49x^{2} to get 11x^{2}.

$$11x^{2}-88x-20=-70x+25$$

Add 70x to both sides.

$$11x^{2}-88x-20+70x=25$$

Combine -88x and 70x to get -18x.

$$11x^{2}-18x-20=25$$

Subtract 25 from both sides.

$$11x^{2}-18x-20-25=0$$

Subtract 25 from -20 to get -45.

$$11x^{2}-18x-45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 11x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=-18 ab=11\left(-45\right)=-495$$

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 -495.

$$1,-495 3,-165 5,-99 9,-55 11,-45 15,-33$$

Calculate the sum for each pair.

$$1-495=-494 3-165=-162 5-99=-94 9-55=-46 11-45=-34 15-33=-18$$

The solution is the pair that gives sum -18.

$$a=-33 b=15$$

Rewrite 11x^{2}-18x-45 as \left(11x^{2}-33x\right)+\left(15x-45\right).

$$\left(11x^{2}-33x\right)+\left(15x-45\right)$$

Factor out 11x in the first and 15 in the second group.

$$11x\left(x-3\right)+15\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(11x+15\right)$$

To find equation solutions, solve x-3=0 and 11x+15=0.

$$x=3 x=-\frac{15}{11}$$

Substitute -\frac{15}{11} for x in the equation \sqrt{5x+1}-\sqrt{3x-5}=\sqrt{x+1}. The expression \sqrt{5\left(-\frac{15}{11}\right)+1} is undefined because the radicand cannot be negative.

$$\sqrt{5\left(-\frac{15}{11}\right)+1}-\sqrt{3\left(-\frac{15}{11}\right)-5}=\sqrt{-\frac{15}{11}+1}$$

Substitute 3 for x in the equation \sqrt{5x+1}-\sqrt{3x-5}=\sqrt{x+1}.

$$\sqrt{5\times 3+1}-\sqrt{3\times 3-5}=\sqrt{3+1}$$

Simplify. The value x=3 satisfies the equation.

$$2=2$$

Equation \sqrt{5x+1}-\sqrt{3x-5}=\sqrt{x+1} has a unique solution.

$$x=3$$