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Zorluk: ZorReal Numbers, Number Line, and Absolute Value

If xx is a real number such that x29=5x3|x^2 - 9| = 5|x - 3|, which of the following could be the value of xx? Select all such values.

  1. 8-8Cevap
  2. B
    2-2
  3. 22Cevap
  4. 33Cevap
  5. E
    88

Cevap

The values of xx that satisfy the equation are 8-8, 22, and 33.
Factoring the left side of x29=5x3|x^2 - 9| = 5|x - 3| yields x3x+3=5x3|x - 3||x + 3| = 5|x - 3|. Setting the common factor x3=0|x - 3| = 0 gives x=3x = 3. Dividing both sides by the non-zero quantity x3|x - 3| leaves x+3=5|x + 3| = 5, which splits into x+3=5    x=2x + 3 = 5 \implies x = 2 and x+3=5    x=8x + 3 = -5 \implies x = -8. Therefore, 8-8, 22, and 33 are all valid solutions.

Adım Adım Çözüm

1
Apply the product rule for absolute values to factor the left-hand side.
x29=(x3)(x+3)=x3x+3|x^2 - 9| = |(x - 3)(x + 3)| = |x - 3| \cdot |x + 3|, rewriting the equation as x3x+3=5x3|x - 3| \cdot |x + 3| = 5|x - 3|.
The absolute value of a product is equal to the product of the individual absolute values.
2
Evaluate the case where the shared factor is zero: x3=0|x - 3| = 0.
x3=0    x=3x - 3 = 0 \implies x = 3. Both sides equal 00, making x=3x = 3 a valid solution.
Dividing by x3|x - 3| without checking if it can be zero would cause the loss of the root x=3x = 3.
3
Evaluate the case where x30|x - 3| \neq 0 by dividing both sides of the equation by x3|x - 3|.
x+3=5|x + 3| = 5.
Since x3>0|x - 3| > 0, we can safely divide both sides by this non-zero quantity.
4
Solve the remaining absolute value equation x+3=5|x + 3| = 5.
x+3=5    x=2x + 3 = 5 \implies x = 2, and x+3=5    x=8x + 3 = -5 \implies x = -8.
An expression inside an absolute value equal to a positive number kk can equal either kk or k-k.

Anahtar Kavram

Factoring absolute value expressions using ab=ab|ab| = |a||b| and systematically considering all cases to avoid dropping zero-roots or negative solutions.
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