Question

If \[\alpha \] and \[{\alpha ^2}\], are the roots \[{x^2} + x + 1 = 0\], then the equation, whose roots are \[{\alpha ^{31}}\] and \[{\alpha ^{62}}\], is
1. ${x^2} - x + 1 = 0$
2. ${x^2} + x - 1 = 0$
3. ${x^2} + x + 1 = 0$
4. ${x^{60}} + {x^{30}} + 1 = 0$

Answer 1

Hint:In this question, we are given the quadratic equation \[{x^2} + x + 1 = 0\] whose roots are \[\alpha \] and \[{\alpha ^2}\]. We have to find the equation whose root are given as \[{\alpha ^{31}}\] and \[{\alpha ^{62}}\]. Apply the formula of the sum of root and product of root using the first quadratic equation. Then, put the required values to form the equation of \[{\alpha ^{31}}\] and \[{\alpha ^{62}}\] roots.

Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Given that,
\[\alpha \] and \[{\alpha ^2}\], are the roots \[{x^2} + x + 1 = 0\]
Compare the quadratic equation \[{x^2} + x + 1 = 0\] with the general equation i.e., \[A{x^2} + Bx + C = 0\]
It implies that,
$A = 1,B = 1,C = 1$
Using the formula of the sum of root and product of root,
We get,
Sum, \[\alpha + {\alpha ^2} = \dfrac{{ - B}}{A} = \dfrac{{ - 1}}{1}\]
$ \Rightarrow \alpha \left( {1 + \alpha } \right) = - 1.................\left( 1 \right)$
Product, \[\alpha {\alpha ^2} = \dfrac{C}{A} = \dfrac{1}{1}\]
$ \Rightarrow {\alpha ^3} = 1.................\left( 2 \right)$
Now, to form a quadratic equation of the given roots \[{\alpha ^{31}}\] and \[{\alpha ^{62}}\]
Here, sum of the roots is \[{\alpha ^{31}} + {\alpha ^{62}} = {\alpha ^{30}}\left( {\alpha + {\alpha ^{32}}} \right)\]
\[{\alpha ^{31}} + {\alpha ^{62}} = {\alpha ^{30}}\left( {\alpha + {\alpha ^{30}}{\alpha ^2}} \right)\]
We can write \[{\alpha ^{30}} = {\left( {{\alpha ^3}} \right)^{10}} = {\left( 1 \right)^{10}} = 1\] (using equation 2)
$ \Rightarrow $ Sum \[ = \left( {\alpha + {\alpha ^2}} \right) = - 1 = \dfrac{{ - B}}{A}\] (From equation 1) -------(3)
Similarly, Product of the roots \[{\alpha ^{31}}\] and \[{\alpha ^{62}}\] will be
\[{\alpha ^{31}} \times {\alpha ^{62}} = {\alpha ^{31 + 62}}\]
\[{\alpha ^{31}} \times {\alpha ^{62}} = {\alpha ^{93}} = {\left( {{\alpha ^3}} \right)^{31}} = 1 = \dfrac{C}{A}\] (From equation 2) --------(4)
Now, using equation (3) and equation (4)
We get $A = 1,B = 1,C = 1$
As we know,
General quadratic equation is \[A{x^2} + Bx + C = 0\]
Put $A = 1,B = 1,C = 1$ in above equation
Thus, the required equation is \[{x^2} + x + 1 = 0\].

Hence, the correct option is (3).

Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.