Question

The number of solutions of the equation \(9x^2 - 18|x| + 5 = 0\) belonging to the domain of definition of \(\log_e \{(x+1)(x+2)\}\), is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Always check domain restrictions after solving the equation.

Answer 1

The Correct Option is C

To find the number of solutions for the equation \(9x^2 - 18|x| + 5 = 0\) that belong to the domain of definition of \(\log_e \{(x+1)(x+2)\}\), we need to follow these steps:

1.  Analyzing the domain of the logarithm:
   The expression inside the logarithm, \((x+1)(x+2)\), must be greater than zero for the logarithm to be defined. Thus, the domain of \(\log_e \{(x+1)(x+2)\}\) is:
   • \((x+1) > 0\) and \((x+2) > 0\), which gives \(x > -1\), or
   • \((x+1) < 0\) and \((x+2) < 0\), which is not possible since both terms cannot be negative simultaneously.
2. Solving the quadratic equation:
   We solve \(9x^2 - 18|x| + 5 = 0\) by considering two cases, depending on the value of \(x\) which affects \(|x|\):
   • For \(x \geq 0\), \(|x| = x\), so the equation becomes: \(9x^2 - 18x + 5 = 0\). Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -18\), \(c = 5\): \(x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 9 \cdot 5}}{18}\), \(x = \frac{18 \pm \sqrt{324 - 180}}{18}\), \(x = \frac{18 \pm \sqrt{144}}{18}\), \(x = \frac{18 \pm 12}{18}\).
     • \(x = \frac{30}{18} = \frac{5}{3}\)
     • \(x = \frac{6}{18} = \frac{1}{3}\)
   • For \(x < 0\), \(|x| = -x\), thus: \(9x^2 + 18x + 5 = 0\). This equation is also solved using the quadratic formula, resulting in: \(x = \frac{-18 \pm \sqrt{18^2 - 4 \times 9 \times 5}}{18}\), \(x = \frac{-18 \pm \sqrt{324 - 180}}{18}\), \(x = \frac{-18 \pm 12}{18}\).
     • \(x = \frac{-6}{18} = -\frac{1}{3}\)
     • \(x = \frac{-30}{18} = -\frac{5}{3}\) (which lies outside our domain filter for \(x > -1\)).

Conclusion: The number of solutions satisfying the original equation and falling within the domain of the logarithmic expression is 3.