Question

If \[a=\cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right)\] then quadratic equation whose roots are \[\alpha =a+{{a}^{2}}+{{a}^{4}}\]and \[\beta ={{a}^{3}}+{{a}^{5}}+{{a}^{6}}\]is
1. \[{{x}^{2}}-x+2=0\]
2. \[{{x}^{2}}+x-2=0\]
3. \[{{x}^{2}}-x-2=0\]
4. \[{{x}^{2}}+x+2=0\]

Answer 1

Hint: In this particular problem we have to find the equation for which we need to use a general quadratic equation that is\[{{x}^{2}}+(\alpha +\beta )x+\alpha \beta =0\]. For that we have to calculate \[(\alpha +\beta )\](sum of roots) and \[\alpha \beta \](products of roots) to solve further and simplifying further we get the required equation.

Complete step-by-step solution:
According to this question it is given the value of \[a=\cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right)\]
As well as two roots that is
\[\alpha =a+{{a}^{2}}+{{a}^{4}}--(1)\]
\[\beta ={{a}^{3}}+{{a}^{5}}+{{a}^{6}}--(2)\]
We have two equation (1) and (2)
First of all we need to find the sum of two equation that means sum of roots
\[\alpha +\beta =(a+{{a}^{2}}+{{a}^{4}})+({{a}^{3}}+{{a}^{5}}+{{a}^{6}})\]
After simplifying and rearranging the term we get:
\[\alpha +\beta =a+{{a}^{2}}+{{a}^{3}}+{{a}^{4}}+{{a}^{5}}+{{a}^{6}}\]
If you notice this above equation then it is looks like sum of a series that means \[{{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}\]
But in the series there are total 6 terms that is \[n=6\]and \[r=a\]
Therefore, substituting this we get:
\[\alpha +\beta =\dfrac{a(1-{{a}^{6}})}{1-a}\]
After simplifying this we get:
\[\alpha +\beta =\dfrac{a-{{a}^{7}}}{1-a}--(3)\]
Now we have to find the value of \[{{a}^{7}}\]that is \[{{a}^{7}}={{\left( \cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right) \right)}^{7}}\]
So we have to use DeMoivre's Theorem that is \[{{\left( \cos (\theta )+i\sin (\theta ) \right)}^{n}}=\cos (n\theta )+i\sin (n\theta )\]
\[{{a}^{7}}=\cos \left( \dfrac{2\pi \times 7}{7} \right)+i\sin \left( \dfrac{2\pi \times 7}{7} \right)\]
That is further simplifying and substituting the \[\cos (2\pi )=1\]and\[\sin (2\pi )=1\]
\[{{a}^{7}}=1\] Substitute this value in equation (3)
\[\alpha +\beta =\dfrac{a-1}{1-a}\]
By simplifying further we get:
\[\alpha +\beta =-1\]
Now, we have to find the product of roots
\[\alpha \beta =(a+{{a}^{2}}+{{a}^{4}})({{a}^{3}}+{{a}^{5}}+{{a}^{6}})\]
Now, by simplifying and further solving we get:
\[\alpha \beta =a(1+a+{{a}^{3}})\times {{a}^{3}}\times (1+{{a}^{2}}+{{a}^{3}})\]
By simplifying further we get:
\[\alpha \beta ={{a}^{4}}(1+a+{{a}^{3}})(1+{{a}^{2}}+{{a}^{3}})\]
Multiply the brackets and simplify the above equation
\[\alpha \beta ={{a}^{4}}(1+{{a}^{2}}+{{a}^{3}}+a+{{a}^{3}}+{{a}^{4}}+{{a}^{3}}+{{a}^{5}}+{{a}^{6}})\]
\[{{a}^{4}}\]Multiply inside the equation which we get:
\[\alpha \beta ={{a}^{4}}+{{a}^{6}}+{{a}^{7}}+{{a}^{5}}+{{a}^{7}}+{{a}^{8}}+{{a}^{7}}+{{a}^{9}}+{{a}^{10}}\]
If you notice in the above equation that \[{{a}^{7}}=1\]substitute in this equation and also further solving we get:
By rearranging the term we get:
\[\alpha \beta =3+({{a}^{4}}+{{a}^{5}}+{{a}^{6}})+({{a}^{8}}+{{a}^{9}}+{{a}^{10}})\]
By simplifying and further solving we get:
\[\alpha \beta =3+({{a}^{4}}+{{a}^{5}}+{{a}^{6}})+({{a}^{7}}.a+{{a}^{7}}.{{a}^{2}}+{{a}^{7}}.{{a}^{3}})\]
Then we have to substitute the value of \[{{a}^{7}}=1\]
\[\alpha \beta =3+({{a}^{4}}+{{a}^{5}}+{{a}^{6}})+(a+{{a}^{2}}+{{a}^{3}})\]
Rearrange the term to get in the form of sum of series.
\[\alpha \beta =3+(a+{{a}^{2}}+{{a}^{3}}+{{a}^{4}}+{{a}^{5}}+{{a}^{6}})\]
Apply the formula of \[{{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}\] But in the series there are total 6 terms that is \[n=6\]and \[r=a\]
Further simplifying we get:
\[\alpha \beta =3+\dfrac{a(1-{{a}^{6}})}{1-a}\]
Further simplifying we get:
\[\alpha \beta =3+\dfrac{a-{{a}^{7}}}{1-a}\]
Substitute \[{{a}^{7}}=1\]in the above equation
\[\alpha \beta =3+\dfrac{a-1}{1-a}\]
By solving further we get:
\[\alpha \beta =3+(-1)\]
\[\alpha \beta =3-1\]
By solving further we get:
\[\alpha \beta =2--(4)\]
General formula for quadratic form is
\[{{x}^{2}}+(\alpha +\beta )x+\alpha \beta =0---(5)\]
We substitute the equation (3) and equation (4) on equation (5) we get:
\[{{x}^{2}}+(-1)x+2=0\]
Hence, we get the required equation is
\[{{x}^{2}}-x+2=0\]
So, the correct option is “option 1”.


Note: In this particular problem we have to keep in mind that the general formula of quadratic and substitute the value to get the required answer don’t make silly mistakes while simplifying the steps because initially we need to calculate the sum and product of roots. Observe the sum of series in that and apply the formula accordingly. So, in this way we get the required equation and the above solution is referred for such types of problems.