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secondary 3 | A Maths
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Hi I have tried many times to solve this question but failed. Appreciate if anyone can help. Thanks in advance
Date Posted: 3 years ago
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Given :
f(x) = 2x² + 3px - 2q
g(x) = x² + q
p,q are constants, p,q ≠ 0
(x - a) is a common factor of both f(x) and g(x) , where a ≠ 0
Since (x - a) is a factor of both functions, then by the Factor Theorem, f(a) = 0 and g(a) = 0
(Remainder = 0 when x = a)
f(a) = 0 → 2a² + 3pa - 2q = 0 ①
g(a) = 0 → a² + q = 0
a² = -q ②
Sub ② into ①,
2(-q) + 3pa - 2q = 0
3pa - 4q = 0
3pa = 4q
The trick here, is to square both sides, so we can get p² on one of them.
(3pa)² = (4q)²
9p²a² = 16q²
16q² - 9p²a² = 0
Since a² = -q,
16q² - 9p²(-q) = 0
16q² + 9p²q = 0
Since q is a non-zero constant, we can reduce the equation by dividing throughout by q .
16q + 9p² = 0
Rewrite as : 9p² + 16q = 0 (shown)
f(x) = 2x² + 3px - 2q
g(x) = x² + q
p,q are constants, p,q ≠ 0
(x - a) is a common factor of both f(x) and g(x) , where a ≠ 0
Since (x - a) is a factor of both functions, then by the Factor Theorem, f(a) = 0 and g(a) = 0
(Remainder = 0 when x = a)
f(a) = 0 → 2a² + 3pa - 2q = 0 ①
g(a) = 0 → a² + q = 0
a² = -q ②
Sub ② into ①,
2(-q) + 3pa - 2q = 0
3pa - 4q = 0
3pa = 4q
The trick here, is to square both sides, so we can get p² on one of them.
(3pa)² = (4q)²
9p²a² = 16q²
16q² - 9p²a² = 0
Since a² = -q,
16q² - 9p²(-q) = 0
16q² + 9p²q = 0
Since q is a non-zero constant, we can reduce the equation by dividing throughout by q .
16q + 9p² = 0
Rewrite as : 9p² + 16q = 0 (shown)
Once again, thank you very much for helping.
Hi thank you once again for answering my question. I have posted a Sec3 IP set language and notation question many days ago but there is no reply. Appreciate if you could help me. Do you want me to type the question or you could search for my qn?
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