Solve for x: $\sqrt{3x + 1} - \sqrt{5x - 1} = \frac{4}{3} - 3x$

Steps to Solve

1. Isolate one of the square roots Add $\sqrt{5x-1}$ to both sides of the equation: $$ \sqrt{3x+1} = \frac{4}{3} - 3x + \sqrt{5x-1} $$

2. Square both sides Squaring both sides to eliminate the first square root: $$ (\sqrt{3x+1})^2 = (\frac{4}{3} - 3x + \sqrt{5x-1})^2 $$ $$ 3x+1 = (\frac{4}{3} - 3x)^2 + 2(\frac{4}{3} - 3x)\sqrt{5x-1} + (\sqrt{5x-1})^2 $$ $$ 3x+1 = \frac{16}{9} - 8x + 9x^2 + 2(\frac{4}{3} - 3x)\sqrt{5x-1} + 5x-1 $$ $$ 3x+1 = 9x^2 - 3x + \frac{7}{9} + 2(\frac{4}{3} - 3x)\sqrt{5x-1} $$

3. Isolate the remaining square root term Move all terms without the square root to the left side: $$ 3x+1 - 9x^2 + 3x - \frac{7}{9} = 2(\frac{4}{3} - 3x)\sqrt{5x-1} $$ $$ -9x^2 + 6x + \frac{2}{9} = 2(\frac{4}{3} - 3x)\sqrt{5x-1} $$ $$ \frac{1}{9}(-81x^2 + 54x + 2) = 2(\frac{4-9x}{3})\sqrt{5x-1} $$ $$ \frac{1}{9}(-81x^2 + 54x + 2) = \frac{2}{3}(4-9x)\sqrt{5x-1} $$

4. Square both sides again Square both sides to eliminate the remaining square root. $$ (\frac{1}{9}(-81x^2 + 54x + 2))^2 = (\frac{2}{3}(4-9x)\sqrt{5x-1})^2 $$ $$ \frac{1}{81}(-81x^2 + 54x + 2)^2 = \frac{4}{9}(4-9x)^2(5x-1) $$ $$ \frac{1}{81}(6561x^4 - 8748x^3 + 2916x^2 + 216x + 4) = \frac{4}{9}(16 - 72x + 81x^2)(5x-1) $$ Multiply both sides by 81: $$ 6561x^4 - 8748x^3 + 2916x^2 + 216x + 4 = 36(16 - 72x + 81x^2)(5x-1) $$ $$ 6561x^4 - 8748x^3 + 2916x^2 + 216x + 4 = 36(405x^3 - 81x^2 - 360x^2 + 72x + 80x - 16) $$ $$ 6561x^4 - 8748x^3 + 2916x^2 + 216x + 4 = 36(405x^3 - 441x^2 + 152x - 16) $$ $$ 6561x^4 - 8748x^3 + 2916x^2 + 216x + 4 = 14580x^3 - 15876x^2 + 5472x - 576 $$

5. Simplify the equation Move all terms to the left side: $$ 6561x^4 - 8748x^3 - 14580x^3 + 2916x^2 + 15876x^2 + 216x - 5472x + 4 + 576 = 0 $$ $$ 6561x^4 - 23328x^3 + 18792x^2 - 5256x + 580 = 0 $$

6. Solve for x This is a quartic equation, which is generally difficult to solve analytically. However, by inspection, we can test simple rational roots. Test $x = \frac{1}{3}$: $$ 6561(\frac{1}{3})^4 - 23328(\frac{1}{3})^3 + 18792(\frac{1}{3})^2 - 5256(\frac{1}{3}) + 580 = 9 - 864 + 2088 - 1752 + 580 = 0 $$ So, $x = \frac{1}{3}$ is a root.

Test $x = \frac{2}{9}$: $$ \sqrt{3(\frac{2}{9}) + 1} - \sqrt{5(\frac{2}{9}) - 1} = \sqrt{\frac{2}{3} + 1} - \sqrt{\frac{10}{9}-1} = \sqrt{\frac{5}{3}} - \sqrt{\frac{1}{9}} = \sqrt{\frac{5}{3}} - \frac{1}{3} $$ $$ \frac{4}{3} - 3(\frac{2}{9}) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} $$ Clearly $x = \frac{2}{9}$ is not a solution.

7. Check for extraneous solutions Substitute $x = \frac{1}{3}$ into the original equation: $$ \sqrt{3(\frac{1}{3}) + 1} - \sqrt{5(\frac{1}{3}) - 1} = \sqrt{1+1} - \sqrt{\frac{5}{3} - 1} = \sqrt{2} - \sqrt{\frac{2}{3}} = \sqrt{2} - \frac{\sqrt{6}}{3} = \sqrt{2}(1-\frac{\sqrt{3}}{3}) $$ $$ \frac{4}{3} - 3(\frac{1}{3}) = \frac{4}{3} - 1 = \frac{1}{3} $$ $$ \sqrt{2} - \sqrt{\frac{2}{3}} = \frac{1}{3} $$ means $ \sqrt{2}(1-\frac{\sqrt{3}}{3}) = \frac{1}{3} $ $$ \sqrt{2}(\frac{3 - \sqrt{3}}{3}) = \frac{1}{3} $$ $$ \sqrt{2}(3 - \sqrt{3}) = 1 $$ $$ \sqrt{2}(3 - \sqrt{3}) \approx 1.79 \ne 1 $$ So, $x = \frac{1}{3}$ is not a valid root. Looks like it is an extraneous solution.

8. Try to manipulate the equation at the start:

Original equation: $\sqrt{3x + 1} - \sqrt{5x - 1} = \frac{4}{3} - 3x$ Let $x = \frac{13}{36}$ $$ \sqrt{3*\frac{13}{36} + 1} - \sqrt{5*\frac{13}{36} - 1} = \sqrt{\frac{13}{12} + \frac{12}{12}} - \sqrt{\frac{65}{36} - \frac{36}{36}} = \sqrt{\frac{25}{12}} - \sqrt{\frac{29}{36}} = \frac{5}{2\sqrt{3}} - \frac{\sqrt{29}}{6} = \frac{5\sqrt{3}}{6} - \frac{\sqrt{29}}{6} \approx 0.6728 $$ $$ \frac{4}{3} - 3*(\frac{13}{36}) = \frac{4}{3} - \frac{13}{12} = \frac{16}{12} - \frac{13}{12} = \frac{3}{12} = \frac{1}{4} = 0.25$$ The original equation is actually quite difficult.

Let's try $x = \frac{1}{5}$ $$ \sqrt{\frac{3}{5} + 1} - \sqrt{1 - 1} = \sqrt{\frac{8}{5}} = \frac{2\sqrt{10}}{5} \approx 1.26 $$ $$ \frac{4}{3} - \frac{3}{5} = \frac{20 - 9}{15} = \frac{11}{15} \approx 0.73 $$ So $x = \frac{1}{5}$ is not a solution

Let's try $ x = \frac{1}{4}$ $$ \sqrt{\frac{3}{4} + 1} - \sqrt{\frac{5}{4} - 1} = \sqrt{\frac{7}{4}} - \sqrt{\frac{1}{4}} = \frac{\sqrt{7}}{2} - \frac{1}{2} \approx 0.823 $$ $$ \frac{4}{3} - \frac{3}{4} = \frac{16 - 9}{12} = \frac{7}{12} \approx 0.583 $$ Looking like there is no rational root