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If the product of two successive positive integers is 7482, which is the greater integer ?
- A87
- B82
- C84
- D89
Answer not set
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0
CCommunity
Solution
Let the two successive positive integers be $(x)$ & $(x + 1)$
Acc to ques,
=> $(x) \times (x + 1) = 7482$
=> $x^2 + x - 7482 = 0$
=> $x = \frac{-1 \pm \sqrt{(1^2) - (4 \times 1 \times -7482)}}{2}$
=> $x = \frac{-1 \pm \sqrt{29929}}{2}$
=> $x = \frac{-1 \pm 173}{2}$
=> $x = 86 , -87$
Since the numbers are positive, => $x = 86$
$\therefore$ Greater number = 86 + 1 = 87
Let the two successive positive integers be $(x)$ & $(x + 1)$
Acc to ques,
=> $(x) \times (x + 1) = 7482$
=> $x^2 + x - 7482 = 0$
=> $x = \frac{-1 \pm \sqrt{(1^2) - (4 \times 1 \times -7482)}}{2}$
=> $x = \frac{-1 \pm \sqrt{29929}}{2}$
=> $x = \frac{-1 \pm 173}{2}$
=> $x = 86 , -87$
Since the numbers are positive, => $x = 86$
$\therefore$ Greater number = 86 + 1 = 87
0
CCommunity
Solution
Let the two successive positive integers be $(x)$ & $(x + 1)$
Acc to ques,
=> $(x) \times (x + 1) = 7482$
=> $x^2 + x - 7482 = 0$
=> $x = \frac{-1 \pm \sqrt{(1^2) - (4 \times 1 \times -7482)}}{2}$
=> $x = \frac{-1 \pm \sqrt{29929}}{2}$
=> $x = \frac{-1 \pm 173}{2}$
=> $x = 86 , -87$
Since the numbers are positive, => $x = 86$
$\therefore$ Greater number = 86 + 1 = 87
Let the two successive positive integers be $(x)$ & $(x + 1)$
Acc to ques,
=> $(x) \times (x + 1) = 7482$
=> $x^2 + x - 7482 = 0$
=> $x = \frac{-1 \pm \sqrt{(1^2) - (4 \times 1 \times -7482)}}{2}$
=> $x = \frac{-1 \pm \sqrt{29929}}{2}$
=> $x = \frac{-1 \pm 173}{2}$
=> $x = 86 , -87$
Since the numbers are positive, => $x = 86$
$\therefore$ Greater number = 86 + 1 = 87