Question:
The interval containing all the solutions of the inequality
\[
\sqrt{x^2-12x+48}\lt 2x+3
\]
only is
The interval containing all the solutions of the inequality
\[
\sqrt{x^2-12x+48}\lt 2x+3
\]
only is
Show Hint
For inequalities involving square roots, first ensure that the right-hand side is positive before squaring. After solving the resulting quadratic inequality, always check the obtained solutions against the original restriction.
Updated On: Jul 29, 2026
- \((-\infty,-\sqrt{29}-4)\cup(\sqrt{29}-4,\infty)\)
- \((-\infty,-\sqrt{29}-4)\)
- \((\sqrt{29}-4,\infty)\)
- \((-\sqrt{29}-4,\sqrt{29}-4)\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For an inequality of the form
\[
\sqrt{A}\lt B,
\]
it is necessary that \(B\gt 0\). Then we can square both sides to obtain
\[
A\lt B^2.
\]
Step 1: Apply the condition \(2x+3\gt 0\). Since \[ \sqrt{x^2-12x+48}\ge 0, \] we must have \[ 2x+3\gt 0. \] Thus, \[ x\gt -\frac{3}{2}. \]
Step 2: Square both sides. Given \[ \sqrt{x^2-12x+48}\lt 2x+3, \] squaring, \[ x^2-12x+48\lt (2x+3)^2. \] \[ x^2-12x+48\lt 4x^2+12x+9. \] \[ 0\lt 3x^2+24x-39. \] \[ x^2+8x-13\gt 0. \]
Step 3: Solve the quadratic inequality. Factor roots using the quadratic formula: \[ x=\frac{-8\pm\sqrt{64+52}}{2} =\frac{-8\pm\sqrt{116}}{2} =-4\pm\sqrt{29}. \] Hence, \[ x^2+8x-13\gt 0 \] gives \[ x\lt -\sqrt{29}-4 \] or \[ x\gt \sqrt{29}-4. \]
Step 4: Apply the restriction \(x\gt -\frac32\). Intersecting \[ (-\infty,-\sqrt{29}-4)\cup(\sqrt{29}-4,\infty) \] with \[ \left(-\frac32,\infty\right), \] the first interval is rejected because \[ -\sqrt{29}-4\lt -\frac32. \] Therefore, the solution set is \[ (\sqrt{29}-4,\infty). \] \[ \boxed{(\sqrt{29}-4,\infty)} \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Apply the condition \(2x+3\gt 0\). Since \[ \sqrt{x^2-12x+48}\ge 0, \] we must have \[ 2x+3\gt 0. \] Thus, \[ x\gt -\frac{3}{2}. \]
Step 2: Square both sides. Given \[ \sqrt{x^2-12x+48}\lt 2x+3, \] squaring, \[ x^2-12x+48\lt (2x+3)^2. \] \[ x^2-12x+48\lt 4x^2+12x+9. \] \[ 0\lt 3x^2+24x-39. \] \[ x^2+8x-13\gt 0. \]
Step 3: Solve the quadratic inequality. Factor roots using the quadratic formula: \[ x=\frac{-8\pm\sqrt{64+52}}{2} =\frac{-8\pm\sqrt{116}}{2} =-4\pm\sqrt{29}. \] Hence, \[ x^2+8x-13\gt 0 \] gives \[ x\lt -\sqrt{29}-4 \] or \[ x\gt \sqrt{29}-4. \]
Step 4: Apply the restriction \(x\gt -\frac32\). Intersecting \[ (-\infty,-\sqrt{29}-4)\cup(\sqrt{29}-4,\infty) \] with \[ \left(-\frac32,\infty\right), \] the first interval is rejected because \[ -\sqrt{29}-4\lt -\frac32. \] Therefore, the solution set is \[ (\sqrt{29}-4,\infty). \] \[ \boxed{(\sqrt{29}-4,\infty)} \] \[ \boxed{\text{Answer = (C)}} \]
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