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High School Mathematics - 2
12.6 Symmetric Function of Roots

Any expression f(a,b) involving two numbers a and b is said to be symmetric if it remains unchanged when a and b are interchanged [i.e f(a,b) = f(b,a)].
Some of the symetric functions of a and b are
a2+b2, a3+b3, ab, a2b2+ab, 1/a+1/b, 1/a2 + 1/b2
The expression f(a,b) = a2-b is not symmetric because
f(b,a) = b2-a not equal to a2-b = f(a,b)
All symmetric functions of a and b can be expressed in terms of two symmetric functions a+b and ab. For instance,
a3 + b3 = (a+b)3-3ab(a+b)
1/a + 1/b = (a+b)/ab
1/a2+1/b2 = [(a+b)2-2ab]/(ab)2
a2b+ab2 = ab(a+b)
We thus observe, without actually solving the quadratic equation ax2+bx+c = 0, a not equal to 0, that
  1. we can find the value of every symmetric function involving the roots of the equation in terms of the coefficients of the equation.
  2. we can find a quadratic equation whose roots are any one of the following pairs of numbers.
a2, b2; a3, b3; a2b

Example: If a, b are the roots of the equation 2x2+3x+7 = 0, then find a2+b2
Solution: Since a,b are the roots of the equation , a+b = -3/2 and ab = 7/2
a2 + b2 = (a+b)a2+b2 = (a+b)2 - 2ab
= (-3/2)2-2(7/2) = -19/4


Directions: Solve the following
Q 1: If a, b are the roots of the equation 2x2+3x+7 = 0, then find 1/a2 + 1/b2
Answer:

Q 2: If a, b are the roots of the equation x2-2x+3 = 0, find the equation whose roots are a+2, b+2
Answer:

Q 3: If a, b are the roots of the equation 2x2+3x+7 = 0, then form the equation whose roots are a2, b2
Answer:

Q 4: If a and b are roots of equation 3x2-4x+1 = 0 find equation whose roots are 3a, 3b
Answer:

Q 5: If the sum of the roots of the equation is 3 and sum of their cubes is 63, find the equation.
Answer:

Q 6: If a, b are the roots of the equation 2x2+3x+7 = 0, then find a3+b3/sup>
Answer:

Q 7: If a, b are the roots of the equation 2x2-5x+7 = 0, find the equation whose roots are 2a+3b, 3a+2b
Answer:

Q 8: If a+b = 1, a22+b2+b2 = 2, find a3+b3
Answer:

Question 9: This question is available to subscribers only!

Question 10: This question is available to subscribers only!


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