Question:
write the given numbers as the sum and product of its zeros respectively 1) root 2 < 1/3
Answer:
Let the numbers are a and b
Given sum of zeroes = √2
=> a + b = √2 .................1
and product of zeroes = 1/3
=> ab = 1/3
Now (a - b)2 = (a + b)2 - 4ab
=> (a - b)22 = (√2)2 - 4 * 1/3
=> (a - b)2 = 2 - 4/3
=> (a - b)2 = (6 - 4)/3
=> (a - b)2 = 2/3
=> (a - b) = √(2/3) ............2
Now add equation 1 and 2, we get
2a = √2 + √(2/3)
=> 2a = √2(1 + 1/√3)
=> 2a = (√2/√3)*(√3 + 1)
=> a = (√2/2√3)*(√3 + 1)
=> a = {(√2 *(√3 + 1)}/{(√2*√2*√3}
=> a = (√3 + 1)/(√2*√3)
=> a = (√3 + 1)/√6
Again suntract equation 1 and 2, we get
2b = √2 - √(2/3)
=> 2b = √2(1 - 1/√3)
=> b = (√2/2√3)*(√3 - 1)
=> b = {(√2 *(√3 - 1)}/{(√2*√2*√3}
=> b = (√3 - 1)/(√2*√3)
=> b = (√3 - 1)/√6
So value of a is (√3 + 1)/√6 and value of b is (√3 - 1)/√6
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