Question

3. If \( R=\frac{\sqrt{x+5}}{x+3} \), for which values of \( x \) will \( R \) be: a) real b) non real c) undefined d) equal to zero

Ask by Fernandez Bryant. in South Africa
Jan 20,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

- \( R \) is real when \( x \geq -5 \) and \( x \neq -3 \). - \( R \) is non-real when \( x < -5 \). - \( R \) is undefined when \( x = -3 \). - \( R \) equals zero when \( x = -5 \).

Solution

To analyze the expression \( R = \frac{\sqrt{x+5}}{x+3} \), we need to consider the conditions under which \( R \) is real, non-real, undefined, or equal to zero. ### a) For \( R \) to be real: 1. The square root \( \sqrt{x+5} \) must be defined, which requires: \[ x + 5 \geq 0 \implies x \geq -5 \] 2. The denominator \( x + 3 \) must not be zero, which requires: \[ x + 3 \neq 0 \implies x \neq -3 \] Combining these conditions, \( R \) is real for: \[ x \geq -5 \quad \text{and} \quad x \neq -3 \] ### b) For \( R \) to be non-real: The expression \( R \) will be non-real if the square root is not defined, which occurs when: \[ x + 5 < 0 \implies x < -5 \] ### c) For \( R \) to be undefined: The expression \( R \) is undefined when the denominator is zero: \[ x + 3 = 0 \implies x = -3 \] ### d) For \( R \) to be equal to zero: The expression \( R \) is equal to zero when the numerator is zero: \[ \sqrt{x+5} = 0 \implies x + 5 = 0 \implies x = -5 \] ### Summary of Results: - a) \( R \) is real for \( x \geq -5 \) and \( x \neq -3 \). - b) \( R \) is non-real for \( x < -5 \). - c) \( R \) is undefined for \( x = -3 \). - d) \( R \) is equal to zero for \( x = -5 \).

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Mind Expander

To determine when \( R = \frac{\sqrt{x+5}}{x+3} \) is real, we need to ensure that the expression under the square root is non-negative. This means \( x + 5 \geq 0 \) or \( x \geq -5 \). Additionally, the denominator \( x + 3 \) cannot be zero, which gives us \( x \neq -3 \). Therefore, \( R \) is real for \( x \in [-5, -3) \cup (-3, \infty) \). For \( R \) to be non-real, the expression under the square root must be negative, hence \( x + 5 < 0 \) or \( x < -5 \). In this case, \( R \) will be non-real for \( x < -5 \). Next, the value of \( R \) is undefined when the denominator is zero. Thus, we solve \( x + 3 = 0 \), leading to \( x = -3 \). At this point, \( R \) is undefined. Finally, \( R \) equals zero when the numerator is zero; this occurs when \( \sqrt{x+5} = 0 \). Therefore, we set \( x + 5 = 0 \), leading to \( x = -5 \). In summary: a) Real for \( x \in [-5, -3) \cup (-3, \infty) \) b) Non-real for \( x < -5 \) c) Undefined at \( x = -3 \) d) Equal to zero at \( x = -5 \)

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