Question

 If the zeroes of the polynomial \[{{x}^{3}}-3{{x}^{2}}+x+1\] are \[a-b,a,a+b\], find a and b.

Answer 1

Hint: In this question, we first need to write the sum, sum of product of two roots at a time, product of the roots in terms of the coefficients of the given cubic polynomial. Then solve the equations formed in terms of a and b to get their respective values.

Complete step by step answer:
CUBIC POLYNOMIAL: A cubic polynomial is a polynomial having degree three.
The general form of a cubic polynomial is given by \[{{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0\]
ZERO OF A POLYNOMIAL: A real number \[\alpha \] is a zero of the polynomial p(x), if and only if \[p\left( \alpha \right)=0\]
Now, let us assume the zeroes of the cubic polynomial as
\[\alpha ,\beta ,\gamma \]
Now, from the cubic polynomial \[{{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0\] we have,
Sum of the roots of the cubic polynomial is given by
\[\alpha +\beta +\gamma =\dfrac{-{{a}_{1}}}{{{a}_{0}}}\]
Now, sum of the product of the two roots at a time is given by
\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{{{a}_{2}}}{{{a}_{0}}}\]
Now, product of all the roots is given by
\[\alpha \beta \gamma =\dfrac{-{{a}_{3}}}{{{a}_{0}}}\]
Now, on comparing the zeroes given in the question to the formulae above we get,
\[\alpha =a-b,\beta =a,\gamma =a+b\]
Now, again on comparing the given cubic polynomial with the general form we get,
\[\Rightarrow {{x}^{3}}-3{{x}^{2}}+x+1=0\]
\[{{a}_{0}}=1,{{a}_{1}}=-3,{{a}_{2}}=1,{{a}_{3}}=1\]
Now, from the sum of the roots formula on substituting the respective values we get,
\[\Rightarrow a-b+a+a+b=\dfrac{-\left( -3 \right)}{1}\]
Now, this can be further written as
\[\Rightarrow a+a+a=3\]
Now, on further simplification we get,
\[\Rightarrow 3a=3\]
Now, on dividing with 3 on both sides we get,
\[\therefore a=1\]
Let us now consider the product of the roots formula and substitute the respective values.
\[\Rightarrow \left( a-b \right)\times a\times \left( a+b \right)=\dfrac{-1}{1}\]
Now, on substituting the value of a this can be further written as
\[\Rightarrow \left( 1-b \right)\times 1\times \left( 1+b \right)=-1\]
Now, on further simplification we get,
\[\Rightarrow 1-{{b}^{2}}=-1\]
Now, on rearranging the terms we get,
\[\Rightarrow {{b}^{2}}=2\]
\[\therefore b=\sqrt{2}\]

Note:
Instead of using the formulae of sum of the roots and product of the roots we can also use the sum of the product of two roots taken at a time and sum of the roots. Both the methods give the same result.
Here, we considered the value of b only positive because as already given that the roots are \[a-b,a,a+b\] we get the same values because the value of b is irrational. That means the roots will be conjugate so either we consider the negative value or positive value the roots will be the same for the given cubic polynomial.