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secondary 3 | A Maths
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I need help with this question. Thank you so much :)
Date Posted: 3 years ago
Views: 173
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An idea
Date Posted:
3 years ago
Or you can just solve them together.
Let 12x² - 5y² = 2p²x - 5y
Substitute (1,p) into the equation,
12(1²) - 5p² = 2p²(1) - 5p
12 - 5p² = 2p² - 5p
7p² - 5p - 12 = 0
(7p - 12)(p + 1) = 0
7p - 12 = 0 or p + 1 = 0
7p = 12 or p = -1
p = 12/7
Sub y = 12/7, x = 1 into 12x² - 5y²
12(1²) - 5(12/7)² = 12 - 720/49 = -132/49 ≠ 7
Sub y = -1, x = 1,
12(1²) - 5(1²) = 12 - 5 = 7
So p = 12/7 is rejected and p = -1
Let 12x² - 5y² = 2p²x - 5y
Substitute (1,p) into the equation,
12(1²) - 5p² = 2p²(1) - 5p
12 - 5p² = 2p² - 5p
7p² - 5p - 12 = 0
(7p - 12)(p + 1) = 0
7p - 12 = 0 or p + 1 = 0
7p = 12 or p = -1
p = 12/7
Sub y = 12/7, x = 1 into 12x² - 5y²
12(1²) - 5(12/7)² = 12 - 720/49 = -132/49 ≠ 7
Sub y = -1, x = 1,
12(1²) - 5(1²) = 12 - 5 = 7
So p = 12/7 is rejected and p = -1
Thank you so much :)
Thank you so much :)
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