Examkin

Question

Difficulty: HardRational and Radical Expressions and Equations

When solving the radical equation 2x2−4x−6=x−3\sqrt{2x^2 - 4x - 6} = x - 3 for all real values of xx, one of the solutions obtained from the squared equation is extraneous. What is the value of this extraneous solution?

  1. -5Answer
  2. B
    3
  3. C
    1
  4. D
    -1/2
  5. E
    5

Answer

The value of the extraneous solution is −5-5.
The correct answer is −5-5. Squaring both sides of the equation 2x2−4x−6=x−3\sqrt{2x^2 - 4x - 6} = x - 3 yields 2x2−4x−6=x2−6x+92x^2 - 4x - 6 = x^2 - 6x + 9, which simplifies to x2+2x−15=0x^2 + 2x - 15 = 0. Factoring this quadratic gives (x+5)(x−3)=0(x + 5)(x - 3) = 0, leading to potential solutions of 33 and −5-5. Substituting −5-5 back into the original equation results in the left side simplifying to 88 and the right side simplifying to −8-8. Since 8≠−88 \neq -8, the value −5-5 is an extraneous solution.

Step-by-Step Solution

1
Square both sides of the original radical equation to eliminate the square root.
2x2−4x−6=(x−3)22x^2 - 4x - 6 = (x - 3)^2
Squaring both sides removes the radical on the left side.
2
Expand the right side and move all terms to the left side to set the quadratic equation to zero.
x2+2x−15=0x^2 + 2x - 15 = 0
Expanding (x−3)2(x - 3)^2 gives x2−6x+9x^2 - 6x + 9. Subtracting this expression from both sides yields the simplified quadratic equation.
3
Factor the quadratic equation to find the potential solutions.
(x+5)(x−3)=0(x + 5)(x - 3) = 0, which gives x=−5x = -5 and x=3x = 3.
The factors of −15-15 that add up to 22 are 55 and −3-3.
4
Substitute each potential solution back into the original equation to check for extraneous roots.
For x=3x = 3, 2(3)2−4(3)−6=3−3\sqrt{2(3)^2 - 4(3) - 6} = 3 - 3 simplifies to 0=00 = 0 (valid). For x=−5x = -5, 2(−5)2−4(−5)−6=−5−3\sqrt{2(-5)^2 - 4(-5) - 6} = -5 - 3 simplifies to 8=−88 = -8 (invalid).
Extraneous solutions satisfy the squared equation but do not satisfy the original radical equation due to the sign difference introduced by squaring.

Key Concept

Solving radical equations and identifying extraneous solutions
Estimated Time:2m 0s
Rate this question

Topics

Rational and Radical Expressions and Equations

More questions from this topic