Solve sqrt{3x+1}+sqrt{4x-3}-sqrt{5x+4}=0

$$\sqrt { 3 x + 1 } + \sqrt { 4 x - 3 } - \sqrt { 5 x + 4 } = 0$$

Solve for x

$$x=1$$

Solution Steps

Subtract \sqrt{4x-3}-\sqrt{5x+4} from both sides of the equation.

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

To find the opposite of \sqrt{4x-3}-\sqrt{5x+4}, find the opposite of each term.

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

The opposite of -\sqrt{5x+4} is \sqrt{5x+4}.

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

Square both sides of the equation.

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

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\sqrt{4x-3}+\sqrt{5x+4}\right)^{2}.

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

Calculate \sqrt{4x-3} to the power of 2 and get 4x-3.

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

Calculate \sqrt{5x+4} to the power of 2 and get 5x+4.

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

Combine 4x and 5x to get 9x.

$$3x+1=9x-3-2\sqrt{4x-3}\sqrt{5x+4}+4$$

Add -3 and 4 to get 1.

$$3x+1=9x+1-2\sqrt{4x-3}\sqrt{5x+4}$$

Subtract 9x+1 from both sides of the equation.

$$3x+1-\left(9x+1\right)=-2\sqrt{4x-3}\sqrt{5x+4}$$

To find the opposite of 9x+1, find the opposite of each term.

$$3x+1-9x-1=-2\sqrt{4x-3}\sqrt{5x+4}$$

Combine 3x and -9x to get -6x.

$$-6x+1-1=-2\sqrt{4x-3}\sqrt{5x+4}$$

Subtract 1 from 1 to get 0.

$$-6x=-2\sqrt{4x-3}\sqrt{5x+4}$$

Square both sides of the equation.

$$\left(-6x\right)^{2}=\left(-2\sqrt{4x-3}\sqrt{5x+4}\right)^{2}$$

Expand \left(-6x\right)^{2}.

$$\left(-6\right)^{2}x^{2}=\left(-2\sqrt{4x-3}\sqrt{5x+4}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$36x^{2}=\left(-2\sqrt{4x-3}\sqrt{5x+4}\right)^{2}$$

Expand \left(-2\sqrt{4x-3}\sqrt{5x+4}\right)^{2}.

$$36x^{2}=\left(-2\right)^{2}\left(\sqrt{4x-3}\right)^{2}\left(\sqrt{5x+4}\right)^{2}$$

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

$$36x^{2}=4\left(\sqrt{4x-3}\right)^{2}\left(\sqrt{5x+4}\right)^{2}$$

Calculate \sqrt{4x-3} to the power of 2 and get 4x-3.

$$36x^{2}=4\left(4x-3\right)\left(\sqrt{5x+4}\right)^{2}$$

Calculate \sqrt{5x+4} to the power of 2 and get 5x+4.

$$36x^{2}=4\left(4x-3\right)\left(5x+4\right)$$

Use the distributive property to multiply 4 by 4x-3.

$$36x^{2}=\left(16x-12\right)\left(5x+4\right)$$

Apply the distributive property by multiplying each term of 16x-12 by each term of 5x+4.

$$36x^{2}=80x^{2}+64x-60x-48$$

Combine 64x and -60x to get 4x.

$$36x^{2}=80x^{2}+4x-48$$

Subtract 80x^{2} from both sides.

$$36x^{2}-80x^{2}=4x-48$$

Combine 36x^{2} and -80x^{2} to get -44x^{2}.

$$-44x^{2}=4x-48$$

Subtract 4x from both sides.

$$-44x^{2}-4x=-48$$

Add 48 to both sides.

$$-44x^{2}-4x+48=0$$

Divide both sides by 4.

$$-11x^{2}-x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -11x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-11\times 12=-132$$

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 -132.

$$1,-132 2,-66 3,-44 4,-33 6,-22 11,-12$$

Calculate the sum for each pair.

$$1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1$$

The solution is the pair that gives sum -1.

$$a=11 b=-12$$

Rewrite -11x^{2}-x+12 as \left(-11x^{2}+11x\right)+\left(-12x+12\right).

$$\left(-11x^{2}+11x\right)+\left(-12x+12\right)$$

Factor out 11x in the first and 12 in the second group.

$$11x\left(-x+1\right)+12\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(11x+12\right)$$

To find equation solutions, solve -x+1=0 and 11x+12=0.

$$x=1 x=-\frac{12}{11}$$

Substitute -\frac{12}{11} for x in the equation \sqrt{3x+1}+\sqrt{4x-3}-\sqrt{5x+4}=0. The expression \sqrt{3\left(-\frac{12}{11}\right)+1} is undefined because the radicand cannot be negative.

$$\sqrt{3\left(-\frac{12}{11}\right)+1}+\sqrt{4\left(-\frac{12}{11}\right)-3}-\sqrt{5\left(-\frac{12}{11}\right)+4}=0$$

Substitute 1 for x in the equation \sqrt{3x+1}+\sqrt{4x-3}-\sqrt{5x+4}=0.

$$\sqrt{3\times 1+1}+\sqrt{4\times 1-3}-\sqrt{5\times 1+4}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Equation \sqrt{3x+1}=\sqrt{5x+4}-\sqrt{4x-3} has a unique solution.

$$x=1$$