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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

Consider the equation \(\mathrm{x}^2+4 \mathrm{x}-\mathrm{n}=0\), where \(\mathrm{n} \in[20,100]\) is a natural number. Then the number of all distinct values of \(n\), for which the given equation has integral roots, is equal to

  1. A 7
  2. B 8
  3. C 6
  4. D 5
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Answer & Solution

Correct Answer

(C) 6

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Detailed explanation

\begin{aligned} & \mathrm{x}^2+4 \mathrm{x}+4=\mathrm{n}+4 \\ & (\mathrm{x}+2)^2=\mathrm{n}+4 \\ & \mathrm{x}=-2 \pm \sqrt{\mathrm{n}+4} \\ & \because 20 \leq \mathrm{n} \leq 100 \\ & \sqrt{24} \leq \sqrt{\mathrm{n}+4} \leq \sqrt{104} \\ & \Rightarrow \sqrt{\mathrm{n}+4}…

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