- #1
Mathsonfire
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- 0
#85
Sum and product of roots of quadratic equation
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- Thread starter Mathsonfire
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In summary, The conversation is about a user posting 4 questions and another user giving them more useful titles. The first user is stuck and asks for help, and the second user asks them to show what they have tried. The first user then shares some equations and the second user provides a summary of the equations and their results. The conversation ends with the second user suggesting a solution, and providing a final summary of the conversation.
#85
- #2
MarkFL
Gold Member
MHB
- 13,288
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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
- 13,288
- 12
Please post what you have so far. :)
- #5
MarkFL
Gold Member
MHB
- 13,288
- 12
We know:
\(\displaystyle c+d=10a\)
\(\displaystyle a+b=10c\)
Hence:
\(\displaystyle a+b+c+d=10(a+c)\)
We also know:
\(\displaystyle c^2-11b=a^2-11d\)
\(\displaystyle 11d-11b=a^2-c^2\)
\(\displaystyle 11(d-b)=(a+c)(a-c)\)
And we obtain by subtraction of the first 2 equations:
\(\displaystyle (d-b)-(a-c)=10(a-c)\)
Or:
\(\displaystyle d-b=11(a-c)\)
Hence:
\(\displaystyle 121(a-c)=(a+c)(a-c)\)
Assuming \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
\(\displaystyle a+c=121\)
Hence:
\(\displaystyle a+b+c+d=10\cdot121=1210\)
\(\displaystyle c+d=10a\)
\(\displaystyle a+b=10c\)
Hence:
\(\displaystyle a+b+c+d=10(a+c)\)
We also know:
\(\displaystyle c^2-11b=a^2-11d\)
\(\displaystyle 11d-11b=a^2-c^2\)
\(\displaystyle 11(d-b)=(a+c)(a-c)\)
And we obtain by subtraction of the first 2 equations:
\(\displaystyle (d-b)-(a-c)=10(a-c)\)
Or:
\(\displaystyle d-b=11(a-c)\)
Hence:
\(\displaystyle 121(a-c)=(a+c)(a-c)\)
Assuming \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:
\(\displaystyle a+c=121\)
Hence:
\(\displaystyle a+b+c+d=10\cdot121=1210\)
FAQ: Sum and product of roots of quadratic equation
What is the formula for finding the sum of roots of a quadratic equation?
The formula for finding the sum of roots of a quadratic equation is -b/a, where a and b are the coefficients of the quadratic equation in standard form (ax^2 + bx + c).
How do you find the product of roots of a quadratic equation?
The product of roots of a quadratic equation can be found by using the formula c/a, where c is the constant term and a is the coefficient of the squared term in the standard form (ax^2 + bx + c).
Can the sum of roots and product of roots of a quadratic equation be equal?
Yes, it is possible for the sum of roots and product of roots of a quadratic equation to be equal. This happens when the quadratic equation has two identical roots, which means the sum of roots is equal to the root squared and the product of roots is also equal to the root squared.
What happens to the sum and product of roots when the quadratic equation has no real solutions?
If the quadratic equation has no real solutions, the sum and product of roots will both be imaginary numbers. This happens when the discriminant (b^2 - 4ac) is negative, indicating that the equation has no real solutions.
How can the sum and product of roots of a quadratic equation be used in solving real-world problems?
The sum and product of roots of a quadratic equation can be used in various real-world problems, such as finding the dimensions of a rectangle with a given perimeter and area, or determining the maximum and minimum values of a quadratic function. They can also be used in physics and engineering problems involving motion and optimization.