Consider the equation $x^2 + 4x - n = 0$, where $n \in [20, 100]$ is a natural number. Then the number of all distinct values of $n$, for which the given equation has integral roots, is equal to
Show Hint
- 7
- 8
- 6
- 9
The Correct Option is C
Approach Solution - 1
2. Solve for $x$: \[ x = -2 \pm \sqrt{n + 4} \]
3. Determine the range of $n$: \[ 20 \leq n \leq 100 \] \[ \sqrt{24} \leq \sqrt{n + 4} \leq \sqrt{104} \] \[ 4.9 \leq \sqrt{n + 4} \leq 10.2 \]
4. Find the integer values of $\sqrt{n + 4}$: \[ \sqrt{n + 4} \in \{5, 6, 7, 8, 9, 10\} \] 5. Calculate the number of distinct values of $n$: \[ \text{Number of distinct values} = 6 \] Therefore, the correct answer is (3) 6.
Approach Solution -2
We need all natural numbers \(n \in [20,100]\) such that the quadratic \(x^2+4x-n=0\) has integral roots.
Concept Used:
For \(ax^2+bx+c=0\) to have integer roots, the discriminant must be a perfect square. Here,
\[ D=b^2-4ac=16+4n=4(4+n). \]
Also the roots are \(x=\dfrac{-4\pm\sqrt{D}}{2}=-2\pm\sqrt{4+n}\).
Step-by-Step Solution:
Step 1: Require \(\sqrt{4+n}\in\mathbb{Z}\). Let \(4+n=k^2\) for some integer \(k\ge 1\). Then
\[ n=k^2-4. \]
Step 2: Impose the range \(20\le n\le 100\):
\[ 20\le k^2-4\le 100 \;\Longrightarrow\; 24\le k^2\le 104. \] \[ k=5,6,7,8,9,10 \quad (\text{since }4^2=16<24,\;11^2=121>104). \]
Step 3: Corresponding \(n\)-values:
\[ n=25-4=21,\;36-4=32,\;49-4=45,\;64-4=60,\;81-4=77,\;100-4=96. \]
Final Computation & Result
There are 6 distinct values of \(n\) in \([20,100]\) for which the equation has integral roots.
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