What's the Error in Solving Radical Equations?

[Schaus]

Homework Statement

√3x - 5 +2 = -3
Underlined is under square root.

√2x - 3 = -x + 3
Underlined is under square root.

Homework Equations

The Attempt at a Solution

√3x - 5 +2 = -3
-2 -2
(√3x - 5)² = (-5)²
3x - 5 = 25
3x = 30
x = 10
Solution says no answer but I got one... I tried 3x - 5 ≠ 0 → x ≠5/3

2nd Question
√2x - 3 = -x + 3
(√2x - 3)² = (-x + 3)²
2x - 3 = (-x + 3)(-x + 3)
2x - 3 = x² - 6x + 9
0 = x² - 8x + 12
0 = (x - 2)(x - 6)
x = 6
x = 2
Solution only says 2 not 6. I tried 2x - 3 ≠ 0 → x ≠ 3/2

Wondering what I did wrong.

 

[Mark44]

Homework Statement

√3x - 5 +2 = -3
Underlined is under square root.

Or equivalently, ##\sqrt{3x - 5} = -5##
This equation has no solution. A square root evaluates to a number that is greater than or equal to zero.

√2x - 3 = -x + 3
Underlined is under square root.

Homework Equations

The Attempt at a Solution

√3x - 5 +2 = -3
-2 -2
(√3x - 5)² = (-5)²
3x - 5 = 25
3x = 30
x = 10
Solution says no answer but I got one... I tried 3x - 5 ≠ 0 → x ≠5/3

2nd Question
√2x - 3 = -x + 3
(√2x - 3)² = (-x + 3)²
2x - 3 = (-x + 3)(-x + 3)
2x - 3 = x² - 6x + 9
0 = x² - 8x + 12
0 = (x - 2)(x - 6)
x = 6
x = 2
Solution only says 2 not 6. I tried 2x - 3 ≠ 0 → x ≠ 3/2

Wondering what I did wrong.

When you square both sides of an equation, you have to check to see if you have introduced extraneous roots. Check both of your solutions in the original equation. I'll bet that 6 is not a solution of the original equation.

 

[Schaus]

Thanks so much! I always forget to check for extraneous roots!

 

[Ray Vickson]

Homework Statement

√3x - 5 +2 = -3
Underlined is under square root.

√2x - 3 = -x + 3
Underlined is under square root.

Homework Equations

The Attempt at a Solution

√3x - 5 +2 = -3
-2 -2
(√3x - 5)² = (-5)²
3x - 5 = 25
3x = 30
x = 10
Solution says no answer but I got one... I tried 3x - 5 ≠ 0 → x ≠5/3

2nd Question
√2x - 3 = -x + 3
(√2x - 3)² = (-x + 3)²
2x - 3 = (-x + 3)(-x + 3)
2x - 3 = x² - 6x + 9
0 = x² - 8x + 12
0 = (x - 2)(x - 6)
x = 6
x = 2
Solution only says 2 not 6. I tried 2x - 3 ≠ 0 → x ≠ 3/2

Wondering what I did wrong.

Instead of underlining and explaining, just use parentheses, like this: √(3x-5) +2 = -3. (However, as already explained, this equation has no (real) solutions.)