Question

The sum of all non-integer roots of the equation $ {x^5} - 6{x^4} + 11{x^3} - 5{x^2} - 3x + 2 = 0 $ is:
A. 6
B. -11
C. -5
D. 3

Answer 1

Hint: Let us consider the given equation as $ f\left( x \right) $ . We can find the sum of all the roots (integer and non-integer) by dividing the negative coefficient of $ {x^4} $ by coefficient of $ {x^5} $ . So first find whether $ f\left( 1 \right) $ is equal to zero or not. If it is zero then 1 is a root which means $ x - 1 $ is a factor of $ f\left( x \right) $ . In the same way find all other integer roots possible for $ f\left( x \right) $ . Add all the integer roots and subtract it from the sum of all roots to give the sum of all non-integer roots of $ f\left( x \right) $ .

Complete step-by-step answer:
We are given to find the sum of all non-integer roots of the equation $ {x^5} - 6{x^4} + 11{x^3} - 5{x^2} - 3x + 2 = 0 $
Let the given equation be $ f\left( x \right) $ .
 $ f\left( x \right) = {x^5} - 6{x^4} + 11{x^3} - 5{x^2} - 3x + 2 $
Sum of all the roots of $ f\left( x \right) $ is $ \dfrac{{ - \left( {coefficient\_of\_{x^4}} \right)}}{{coefficient\_of\_{x^5}}} = \dfrac{{ - \left( { - 6} \right)}}{1} = 6 $
When we substitute x=1, we get
 $ f\left( 1 \right) = {1^5} - 6\left( {{1^4}} \right) + 11\left( {{1^3}} \right) - 5\left( {{1^2}} \right) - 3\left( 1 \right) + 2 = 1 - 6 + 11 - 5 - 3 + 2 = 14 - 14 = 0 $
As we can see $ f\left( 1 \right) = 0 $ , so 1 is a root of $ f\left( x \right) $
When we substitute x=2, we get
 $ f\left( 2 \right) = {2^5} - 6\left( {{2^4}} \right) + 11\left( {{2^3}} \right) - 5\left( {{2^2}} \right) - 3\left( 2 \right) + 2 $
 $ \Rightarrow f\left( 2 \right) = 32 - 96 + 88 - 20 - 6 + 2 = 122 - 122 = 0 $
As we can see $ f\left( 2 \right) = 0 $ , so 2 is also a root of $ f\left( x \right) $
Now $ f\left( x \right) $ can be written as $ \left( {x - 1} \right)\left( {x - 2} \right)\left( {{x^3} - 3{x^2} + 1} \right) $
The expression $ \left( {{x^3} - 3{x^2} + 1} \right) $ has non-integer roots which are imaginary and fractions.
Therefore the integer roots of $ f\left( x \right) $ are 1 and 2.
Sum of integer roots is $ 1 + 2 = 3 $
Therefore Sum of non-integer roots is Sum of all the roots minus sum of integer roots.
 $ \Rightarrow 6 - 3 = 3 $
Hence, the correct option is Option D, sum of all non-integer roots of the equation $ {x^5} - 6{x^4} + 11{x^3} - 5{x^2} - 3x + 2 = 0 $ is $3$.
So, the correct answer is “Option D”.

Note: Another approach.
We have got that $ f\left( x \right) $ can also be written as $ \left( {x - 1} \right)\left( {x - 2} \right)\left( {{x^3} - 3{x^2} + 1} \right) $
And $ \left( {{x^3} - 3{x^2} + 1} \right) $ has non-integer roots which are imaginary and fractions.
The sum of all the non-integer roots of $ f\left( x \right) $ is equal to the sum of all the roots of $ \left( {{x^3} - 3{x^2} + 1} \right) $
Sum of roots of $ \left( {{x^3} - 3{x^2} + 1} \right) $ is $ \dfrac{{ - \left( {coefficient\_of\_{x^2}} \right)}}{{coefficient\_of\_{x^3}}} = \dfrac{{ - \left( { - 3} \right)}}{1} = 3 $
Therefore, the sum of all the non-integer roots of $ f\left( x \right) $ is 3.