What's the Error in Solving Radical Equations?

In summary, the given conversation discusses two equations involving square roots and attempts to solve them. However, both equations do not have real solutions. The first equation simplifies to √(3x-5) = -5, which has no solution since the square root cannot be negative. The second equation simplifies to x^2 - 8x + 12 = 0, which has two solutions, x = 2 and x = 6. However, when these solutions are checked in the original equation, it is found that x = 6 is an extraneous root and therefore the only real solution is x = 2.
  • #1
Schaus
118
5

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.
 
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  • #2
Schaus said:

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.
Schaus said:
√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.
 
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  • #3
Thanks so much! I always forget to check for extraneous roots!
 
  • #4
Schaus said:

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.)
 
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Related to What's the Error in Solving Radical Equations?

What is a radical equation?

A radical equation is an equation that contains a variable under a radical sign, such as a square root or cube root.

How do you solve a radical equation?

To solve a radical equation, you must isolate the variable by eliminating the radical. This is usually done by squaring or cubing both sides of the equation, depending on the type of radical.

What are the common mistakes made when solving radical equations?

Some common mistakes when solving radical equations include forgetting to check for extraneous solutions, not simplifying the radical completely, and incorrectly applying the rules of exponents.

What is an extraneous solution?

An extraneous solution is a solution that does not satisfy the original equation, but appears as a solution after solving a radical equation. These solutions must be checked and eliminated if necessary.

Are there any special cases when solving radical equations?

Yes, there are special cases such as when the radical is in the denominator or when there are multiple radicals in the equation. These cases require additional steps to solve the equation correctly.

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