[tex]\sqrt{3x+5} - \sqrt{2x+1} =\sqrt{x+4}[/tex]

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вопрос

To solve this equation, we first square both sides to eliminate the square root:

[tex](\sqrt{3x+5} - \sqrt{2x+1})^2 = (\sqrt{x+4})^2[/tex]

Expanding the left side using the formula (a-b)^2 = a^2 - 2ab + b^2, we get:

[tex]3x + 5 - 2\sqrt{(3x+5)(2x+1)} + 2x + 1 = x + 4[/tex]

Simplify the expression to get:

[tex]5x + 6 - 2\sqrt{6x^2 + 13x + 5} = x + 4[/tex]

Now, isolate the square root term:

[tex]2\sqrt{6x^2 + 13x + 5} = 4x + 2[/tex]

Square both sides again to eliminate the square root:

[tex]4(6x^2 + 13x + 5) = (4x + 2)^2[/tex]

Expanding both sides:

[tex]24x^2 + 52x + 20 = 16x^2 + 16x + 4[/tex]

Simplify the equation and set it equal to 0:

[tex]8x^2 + 36x + 16 = 0[/tex]

Now, we can solve this quadratic equation for x using the quadratic formula:

[tex]x = \frac{-36 \pm \sqrt{(-36)^2 - 4(8)(16)}}{2(8)}[/tex]

[tex]x = \frac{-36 \pm \sqrt{1296 - 512}}{16}[/tex]

[tex]x = \frac{-36 \pm \sqrt{784}}{16}[/tex]

[tex]x = \frac{-36 \pm 28}{16}[/tex]

Therefore, the solutions for the equation are:
[tex]x_1 = \frac{-36 + 28}{16} = \frac{-8}{16} = -0.5[/tex]
[tex]x_2 = \frac{-36 - 28}{16} = \frac{-64}{16} = -4[/tex]