- #1
Mathsonfire
- 11
- 0
#85
Sum and product of roots of quadratic equation
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- Thread starter Mathsonfire
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Discussion Overview
The discussion revolves around the sum and product of the roots of quadratic equations, focusing on deriving relationships between coefficients and roots. Participants are exploring mathematical reasoning related to specific equations and their properties.
Discussion Character
- Mathematical reasoning
Main Points Raised
- One participant expresses difficulty in progressing after finding the sum of the roots, indicating a need for further guidance.
- Another participant requests that the original poster share their attempts to better understand their challenges.
- A detailed mathematical derivation is presented, showing relationships between the coefficients and roots, including equations that relate sums and differences of the roots.
- The derivation leads to a conclusion about the sum of the roots being equal to 1210 under certain assumptions.
Areas of Agreement / Disagreement
The discussion does not show clear consensus, as participants are at different stages of understanding and problem-solving, with some expressing confusion while others provide mathematical insights.
Contextual Notes
The discussion includes assumptions about the distinctness of roots and the relationships between coefficients that may not be universally applicable without further clarification.
#85
- #2
MarkFL
Gold Member
MHB
- 13,284
- 12
You've posted 4 questions, and I have given all 4 threads more useful titles, now I would ask that you show what you've tried in each thread. This way we can see where you're stuck.
- #3
Mathsonfire
- 11
- 0
Well i am stuck at this point
I found sum of zeroes in both cases and then got stuck.what should i do
I found sum of zeroes in both cases and then got stuck.what should i do
- #4
MarkFL
Gold Member
MHB
- 13,284
- 12
Please post what you have so far. :)
- #5
MarkFL
Gold Member
MHB
- 13,284
- 12
We know:
$$c+d=10a$$
$$a+b=10c$$
Hence:
$$a+b+c+d=10(a+c)$$
We also know:
$$c^2-11b=a^2-11d$$
$$11d-11b=a^2-c^2$$
$$11(d-b)=(a+c)(a-c)$$
And we obtain by subtraction of the first 2 equations:
$$(d-b)-(a-c)=10(a-c)$$
Or:
$$d-b=11(a-c)$$
Hence:
$$121(a-c)=(a+c)(a-c)$$
Assuming \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
$$a+c=121$$
Hence:
$$a+b+c+d=10\cdot121=1210$$
$$c+d=10a$$
$$a+b=10c$$
Hence:
$$a+b+c+d=10(a+c)$$
We also know:
$$c^2-11b=a^2-11d$$
$$11d-11b=a^2-c^2$$
$$11(d-b)=(a+c)(a-c)$$
And we obtain by subtraction of the first 2 equations:
$$(d-b)-(a-c)=10(a-c)$$
Or:
$$d-b=11(a-c)$$
Hence:
$$121(a-c)=(a+c)(a-c)$$
Assuming \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
$$a+c=121$$
Hence:
$$a+b+c+d=10\cdot121=1210$$
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