How do you solve and find the extraneous solutions for #2\sqrt{4-3x}+3=0#?

(Notice that it is now clear that this equation has no real solution since the left-hand side cannot be negative.)

by squaring,

which is its extraneous solution.

I hope that this was helpful.

To solve the equation 2√(4-3x) + 3 = 0 and find any extraneous solutions, follow these steps:

1. Subtract 3 from both sides of the equation: 2√(4-3x) = -3.

2. Divide both sides by 2: √(4-3x) = -3/2.

3. Square both sides of the equation to eliminate the square root: 4-3x = (3/2)^2.

4. Simplify the right side: 4-3x = 9/4.

5. Move the constant term to the other side: -3x = 9/4 - 4.

6. Simplify the right side: -3x = 9/4 - 16/4.

7. Combine the fractions: -3x = -7/4.

8. Divide both sides by -3 to solve for x: x = (-7/4) / -3.

9. Simplify the right side: x = 7/12.

Therefore, the solution to the equation is x = 7/12. There are no extraneous solutions in this case.

To solve the equation ( 2\sqrt{4-3x} + 3 = 0 ) and find any extraneous solutions, follow these steps:

1. Start by isolating the square root term on one side of the equation: ( 2\sqrt{4-3x} = -3 )

2. Square both sides of the equation to eliminate the square root: ( (2\sqrt{4-3x})^2 = (-3)^2 ) ( 4(4-3x) = 9 )

3. Expand and simplify the equation: ( 16 - 12x = 9 )

4. Move all the terms involving x to one side of the equation: ( -12x = 9 - 16 ) ( -12x = -7 )

5. Divide both sides by -12 to solve for x: ( x = \frac{-7}{-12} ) ( x = \frac{7}{12} )

6. Check the solution for extraneous solutions by substituting it back into the original equation: ( 2\sqrt{4-3(\frac{7}{12})} + 3 = 0 ) ( 2\sqrt{4 - \frac{7}{4}} + 3 = 0 ) ( 2\sqrt{\frac{16}{4} - \frac{7}{4}} + 3 = 0 ) ( 2\sqrt{\frac{9}{4}} + 3 = 0 ) ( 2(\frac{3}{2}) + 3 = 0 ) ( 3 + 3 = 0 ) ( 6 \neq 0 )

Since the substituted solution ( x = \frac{7}{12} ) does not satisfy the original equation ( 2\sqrt{4-3x} + 3 = 0 ), there are no real solutions to the original equation, and there are no extraneous solutions to consider.