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x^2 + px + 32 = 0
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25 Sep 2021, 19:04
25 Sep 2021, 19:04
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Question Stats:
60% (00:53) correct
40% (01:20) wrong based on 10 sessions
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If the solutions to the expression \(x^2 + px + 32 = 0\) are m and n, where m, n and p are integers. What is the absolute maximum difference between m and n?
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31
Re: x^2 + px + 32 = 0
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26 Sep 2021, 05:40
26 Sep 2021, 05:40
We get two equations with the help of m and n but I can't find any way to solve this.
(x-m)(x-n)=0
OR
m^2 + pm +32 =0
n^2 + pn + 32 =0
How can we solve using these?
(x-m)(x-n)=0
OR
m^2 + pm +32 =0
n^2 + pn + 32 =0
How can we solve using these?
Re: x^2 + px + 32 = 0
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17 Oct 2021, 03:00
17 Oct 2021, 03:00
For max difference m and n should have the greatest difference between them and we know that mn=32, so we have m and n as 32 and 1, and their difference is 31
