Consider the equation , where n [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 [2025]

Question

Consider the equation $x^{2} + 4 x - n = 0$ , 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 [2025]

Smart Achievers · Accepted Answer

(3)

we have, $x^{2} + 4 x - n = 0$

$\Rightarrow x^{2} + 4 x + 4 = n + 4$

$\Rightarrow {(x + 2)}^{2} = n + 4$

$\Rightarrow x = - 2 \pm \sqrt{n + 4}$

But $20 \leq n \leq 100$

$\Rightarrow 24 \leq n + 4 \leq 104 \Rightarrow \sqrt{24} \leq \sqrt{n + 4} \leq \sqrt{104}$

$\Rightarrow \sqrt{n + 4} \in {5, 6, 7, 8, 9, 10}$ .

$∴$ 6 integral values of 'n' are possible.

Solution

Q.
Consider the equation $x^{2} + 4 x - n = 0$ , 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 [2025]

1 5

2 7

3 6

4 8

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Solution

Q. Consider the equation x2+4x–n=0, where n ∈ [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 [2025]

1 5

2 7

3 6

4 8

Ans. (3) we have, x2+4x–n=0 ⇒ x2+4x+4=n+4 ⇒ (x+2)2=n+4 ⇒ x=–2±n+4 But 20≤n≤100 ⇒ 24≤n+4≤104 ⇒ 24≤n+4≤104 ⇒ n+4∈{5,6,7,8,9,10}. ∴ 6 integral values of 'n' are possible.

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Q.
Consider the equation $x^{2} + 4 x - n = 0$ , 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 [2025]