Question

$\sf \: If \alpha \: and \: \beta \: are \: roots \: of \: {x}^{2} + x + 1$

then form an equation whose roots are

${ \alpha }^{14} + { \beta }^{11} \: and \: { \alpha }^{11} + { \beta }^{14}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm \: {x}^{2} + x + 1 = 0$

Let's first evaluate the roots of the equation using Quadratic formula.

So, By using quadratic formula, we have

$\boxed{\sf{ \: x \: = \: \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} \: }}$

So, here

$\rm \: a = 1$

$\rm \: b = 1$

$\rm \: c = 1$

So, on substituting the values, we get

$\rm \: x \: = \: \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\rm \: x \: = \: \dfrac{ - 1 \: \pm \: \sqrt{ 1 - 4} }{2}$

$\rm \: x \: = \: \dfrac{ - 1 \: \pm \: \sqrt{-3} }{2}$

$\rm \: x \: = \: \dfrac{ - 1 \: \pm \: i \: \sqrt{3} }{2}$

$\rm \: x \: = \: \dfrac{ - 1 \: + \: i \: \sqrt{3} }{2} \: \: or \: \: \: \dfrac{ - 1 \: - \: i \: \sqrt{3} }{2}$

We know,

$\rm \: \omega \: = \: \: \dfrac{ - 1 \: + \: i \: \sqrt{3} }{2}$

and

$\rm \: {\omega }^{2} = \: \dfrac{ - 1 \: - \: i \: \sqrt{3} }{2}$

So, Roots of the given quadratic equation are

$\rm \: \alpha = \omega$

and

$\rm \: \beta = {\omega }^{2}$

Now, Consider,

$\rm \: { \alpha }^{14} + { \beta }^{11}$

$\rm \: = \: {\omega }^{14} + {( {\omega }^{2}) }^{11}$

$\rm \: = \: {\omega }^{2} + {\omega }^{22} \: \: \: \: \: \: \: \{ \because \: {\omega }^{3} = 1 \}$

$\rm \: = \: {\omega }^{2} + \omega \: \: \: \: \: \: \: \{ \because \: {\omega }^{3} = 1 \}$

$\rm \: = \: - \: 1 \: \: \: \: \: \: \: \{ \because \: {1 + \omega + \omega }^{2} = 0 \}$

Now, Consider

$\rm \: { \alpha }^{11} + { \beta }^{14}$

$\rm \: = \: {\omega }^{11} + {( {\omega }^{2}) }^{14}$

$\rm \: = \: {\omega }^{2} + {\omega }^{28} \: \: \: \: \: \: \: \{ \because \: {\omega }^{3} = 1 \}$

$\rm \: = \: {\omega }^{2} + \omega \: \: \: \: \: \: \: \{ \because \: {\omega }^{3} = 1 \}$

$\rm \: = \: - \: 1 \: \: \: \: \: \: \: \{ \because \: {1 + \omega + \omega }^{2} = 0 \}$

So, it means we have to find a quadratic equation whose roots are - 1 and - 1.

$\rm \: Sum \: of \: the \: roots,S = - 1 + ( - 1) = - 2$

and

$\rm \: Product \: of \: the \: roots,P = - 1 \times ( - 1) = 1$

So, the required Quadratic equation is

$\rm \: {x}^{2} - Sx + P = 0$

So, on substituting the values, we get

$\rm \: {x}^{2} + 2x + 1 = 0$

$\rm\implies \: {(x + 1)}^{2} = 0 \: is \: required \: quadratic \: equation.$

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ADDITIONAL INFORMATION:-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac