Find integral values of x for which x2 + 9x + 1 is a perfect square of an integer.
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x = 15 for which x² + 9x + 1 is a perfect square of an integer.
Step-by-step explanation:
x² + 9x + 1 = a²
=> x² + 9x + 1 - a² = 0
x = (- 9 ± √(81 - 4 + 4a²) )/2
x = (- 9 ± √(77 + 4a²) )/2
77 + 4a² = b²
=> b² - 4a² = 77
=> (b + 2a)(b - 2a) = 77
77 = 7 *11 or 1 * 77
b + 2a = 11
b - 2a = 7
=> b = 9 , a = 1
x = (- 9 ± 9)/2 = 0 or - 9
b + 2a = 77
b - 2a = 1
=> b = 39 a = 19
x = (- 9 ± 39)/2 = 15 or - 24
for x = - 24 , - 9 , 0 , 15 x² + 9x + 1 is a perfect square
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