- #1
Mathsonfire
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#85
Sum and product of roots of quadratic equation
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SUMMARY
The discussion focuses on deriving the sum and product of the roots of a quadratic equation using specific relationships between the coefficients. The equations presented include \(c + d = 10a\) and \(a + b = 10c\), leading to the conclusion that \(a + b + c + d = 1210\). The analysis also reveals that \(a + c = 121\) under the assumption that \(a \neq c\). This establishes a clear method for solving quadratic equations based on the relationships between their roots.
PREREQUISITES- Understanding of quadratic equations and their roots
- Familiarity with algebraic manipulation and equations
- Knowledge of the relationships between coefficients in quadratic equations
- Basic proficiency in mathematical notation and expressions
- Study the derivation of the quadratic formula and its applications
- Explore Vieta's formulas for relationships between roots and coefficients
- Learn about the implications of the discriminant in quadratic equations
- Investigate advanced algebraic techniques for solving polynomial equations
Students studying algebra, mathematicians interested in polynomial equations, and educators teaching quadratic equations and their properties.
#85
- #2
MarkFL
Gold Member
MHB
- 13,284
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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
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Please post what you have so far. :)
- #5
MarkFL
Gold Member
MHB
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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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