Solve the equation 2\[0.5ex] vert 2x−5 − 4x−3 = 1

$$(2\sqrt{2x-5})^2 = (1 + \sqrt{4x-3})^2$$ $$4(2x-5) = 1^2 + 2(1)\sqrt{4x-3} + (\sqrt{4x-3})^2$$ $$8x - 20 = 1 + 2\sqrt{4x-3} + 4x - 3$$ $$8x - 20 = 4x - 2 + 2\sqrt{4x-3}$$

$$8x - 20 - (4x - 2) = 2\sqrt{4x-3}$$ $$4x - 18 = 2\sqrt{4x-3}$$

Divide both sides by 2:

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

$$(2x - 9)^2 = (\sqrt{4x-3})^2$$ $$4x^2 - 36x + 81 = 4x - 3$$

$$4x^2 - 36x - 4x + 81 + 3 = 0$$ $$4x^2 - 40x + 84 = 0$$

Divide the entire equation by 4 to simplify:

$$x^2 - 10x + 21 = 0$$

We need two numbers that multiply to 21 and add to -10. These numbers are -3 and -7.

$$(x - 3)(x - 7) = 0$$

This gives two potential solutions: $x = 3$ or $x = 7$.

For $x = 3$:

$$2\sqrt{2(3)-5} - \sqrt{4(3)-3} = 1$$ $$2\sqrt{6-5} - \sqrt{12-3} = 1$$ $$2\sqrt{1} - \sqrt{9} = 1$$ $$2(1) - 3 = 1$$ $$2 - 3 = 1$$ $$-1 = 1$$

This statement is false, so $x=3$ is an extraneous solution.

For $x = 7$:

$$2\sqrt{2(7)-5} - \sqrt{4(7)-3} = 1$$ $$2\sqrt{14-5} - \sqrt{28-3} = 1$$ $$2\sqrt{9} - \sqrt{25} = 1$$ $$2(3) - 5 = 1$$ $$6 - 5 = 1$$ $$1 = 1$$

This statement is true, so $x=7$ is the correct solution.