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Find a quadratic equation whose roots x1 and x2 satisfy the condition
(i) 3(x1^5 + x2^5) = 11(x1^3 + x2^3)
(ii) (x1^2 + x2^2) = 5
(Assume that x1, x2 are real)?
(i) 3(x1^5 + x2^5) = 11(x1^3 + x2^3)
(ii) (x1^2 + x2^2) = 5
(Assume that x1, x2 are real)?
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Find a quadratic equation whose roots x1 and x2 satisfy the condition(...
Understanding the Conditions
To find a quadratic equation with roots x1 and x2, we need to use the given conditions:
- Condition (i): 3(x1^5 + x2^5) = 11(x1^3 + x2^3)
- Condition (ii): x1^2 + x2^2 = 5
Using Vieta's Formulas
Let’s denote:
- s1 = x1 + x2 (sum of roots)
- s2 = x1 * x2 (product of roots)
From Condition (ii), we know:
- x1^2 + x2^2 = (x1 + x2)^2 - 2(x1 * x2) = s1^2 - 2s2 = 5
This gives us our first equation:
- s1^2 - 2s2 = 5 (Equation 1)
Expressing Higher Powers
Now, we need to express x1^3 and x2^3 in terms of s1 and s2:
- x1^3 + x2^3 = (x1 + x2)(x1^2 + x2^2 - x1 * x2) = s1(5 - s2)
Next, for x1^5 + x2^5:
- x1^5 + x2^5 = (x1 + x2)(x1^4 + x2^4) = s1(x1^4 + x2^4)
- Using x1^4 + x2^4 = (x1^2 + x2^2)^2 - 2(x1 * x2)^2 = 5^2 - 2s2^2 = 25 - 2s2^2
This results in:
- x1^5 + x2^5 = s1(25 - 2s2^2)
Substituting into the Given Condition
Now substitute these into Condition (i):
3[s1(25 - 2s2^2)] = 11[s1(5 - s2)]
This simplifies to:
75s1 - 6s1s2^2 = 55s1 - 11s2s1
Cancelling s1 (assuming s1 ≠ 0):
20 = 6s2^2 - 11s2
Rearranging gives a quadratic in s2:
6s2^2 - 11s2 - 20 = 0
Solving the Quadratic
Using the quadratic formula:
s2 = [11 ± sqrt(121 + 480)] / 12
This results in roots for s2, allowing us to find s1. Finally, plug s1 and s2 back into the quadratic equation:
x^2 - s1x + s2 = 0.
This gives the desired quadratic equation with roots x1 and x2.
To find a quadratic equation with roots x1 and x2, we need to use the given conditions:
- Condition (i): 3(x1^5 + x2^5) = 11(x1^3 + x2^3)
- Condition (ii): x1^2 + x2^2 = 5
Using Vieta's Formulas
Let’s denote:
- s1 = x1 + x2 (sum of roots)
- s2 = x1 * x2 (product of roots)
From Condition (ii), we know:
- x1^2 + x2^2 = (x1 + x2)^2 - 2(x1 * x2) = s1^2 - 2s2 = 5
This gives us our first equation:
- s1^2 - 2s2 = 5 (Equation 1)
Expressing Higher Powers
Now, we need to express x1^3 and x2^3 in terms of s1 and s2:
- x1^3 + x2^3 = (x1 + x2)(x1^2 + x2^2 - x1 * x2) = s1(5 - s2)
Next, for x1^5 + x2^5:
- x1^5 + x2^5 = (x1 + x2)(x1^4 + x2^4) = s1(x1^4 + x2^4)
- Using x1^4 + x2^4 = (x1^2 + x2^2)^2 - 2(x1 * x2)^2 = 5^2 - 2s2^2 = 25 - 2s2^2
This results in:
- x1^5 + x2^5 = s1(25 - 2s2^2)
Substituting into the Given Condition
Now substitute these into Condition (i):
3[s1(25 - 2s2^2)] = 11[s1(5 - s2)]
This simplifies to:
75s1 - 6s1s2^2 = 55s1 - 11s2s1
Cancelling s1 (assuming s1 ≠ 0):
20 = 6s2^2 - 11s2
Rearranging gives a quadratic in s2:
6s2^2 - 11s2 - 20 = 0
Solving the Quadratic
Using the quadratic formula:
s2 = [11 ± sqrt(121 + 480)] / 12
This results in roots for s2, allowing us to find s1. Finally, plug s1 and s2 back into the quadratic equation:
x^2 - s1x + s2 = 0.
This gives the desired quadratic equation with roots x1 and x2.
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Find a quadratic equation whose roots x1 and x2 satisfy the condition(i) 3(x1^5 + x2^5) = 11(x1^3 + x2^3)(ii) (x1^2 + x2^2) = 5(Assume that x1, x2 are real)?
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