# If α,β are the roots of x²-x+1=0,then the quadratic equation whose roots are α²⁰¹⁵,β²⁰¹⁵ is ?

${x}^{2} - x + 1 = 0$

#### Explanation:

The roots \alpha\ \ &\ \ \beta of quadratic equation: ${x}^{2} - x + 1 = 0$ are given as

$\setminus \alpha , \setminus \beta = \setminus \frac{- \left(- 1\right) \setminus \pm \setminus \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)}$

$= \setminus \frac{1 \setminus \pm i \setminus \sqrt{3}}{2}$

Let $\setminus \alpha = \setminus \frac{1 + i \setminus \sqrt{3}}{2}$

$\setminus \alpha = {e}^{i \setminus \frac{\pi}{3}}$

$\setminus \therefore \setminus {\alpha}^{2015} = {\left({e}^{i \setminus \frac{\pi}{3}}\right)}^{2015}$

$= {e}^{i \frac{2015 \setminus \pi}{3}}$

$= \setminus \cos \left(\frac{2015 \setminus \pi}{3}\right) + i \setminus \sin \left(\frac{2015 \setminus \pi}{3}\right)$

$= \setminus \cos \left(\setminus \frac{\pi}{3}\right) - i \setminus \sin \left(\setminus \frac{\pi}{3}\right)$

$= \setminus \frac{1 - i \setminus \sqrt{3}}{2}$

Let $\beta = \setminus \frac{1 - i \setminus \sqrt{3}}{2}$

$\setminus \beta = {e}^{- i \setminus \frac{\pi}{3}}$

$\setminus \therefore \setminus {\beta}^{2015} = {\left({e}^{- i \setminus \frac{\pi}{3}}\right)}^{2015}$

$= {e}^{- i \frac{2015 \setminus \pi}{3}}$

$= \setminus \cos \left(- \frac{2015 \setminus \pi}{3}\right) + i \setminus \sin \left(- \frac{2015 \setminus \pi}{3}\right)$

$= \setminus \cos \left(\setminus \frac{\pi}{3}\right) + i \setminus \sin \left(\setminus \frac{\pi}{3}\right)$

$= \setminus \frac{1 + i \setminus \sqrt{3}}{2}$

Since, the roots of new quadratic equation are same as that of given equation ${x}^{2} - x + 1 = 0$ hence new quadratic equation

${x}^{2} - x + 1 = 0$

Jul 28, 2018

If $\alpha \mathmr{and} \beta$ are two different roots of the quadratic equation ${x}^{2} - x + 1 = 0$ then

$\alpha + \beta = 1 \mathmr{and} \alpha \beta = 1$

Also
${\alpha}^{2} - \alpha + 1 = 0 \mathmr{and} {\beta}^{2} - \beta + 1 = 0$

So
${\alpha}^{3} = {\alpha}^{3} + 1 - 1 = \left(\alpha + 1\right) \left({\alpha}^{2} - \alpha + 1\right) - 1$
$= \left(\alpha + 1\right) \times 0 - 1 = - 1$

Similarly ${\beta}^{3} = - 1$

Now ${\alpha}^{2015} = {\alpha}^{3 \times 671 + 2} = {\left({\alpha}^{3}\right)}^{671} {\alpha}^{2} = {\left(- 1\right)}^{871} {\alpha}^{2} = - {\alpha}^{2}$

similarly ${\beta}^{2015} = - {\beta}^{2}$

So sum of the roots of new equation

${\alpha}^{2015} + {\beta}^{2015}$

$= - {\alpha}^{2} - {\beta}^{2}$

$= - \left({\alpha}^{2} + {\beta}^{2}\right)$

$= - \left({\left(\alpha + \beta\right)}^{2} - 2 \alpha \beta\right) = - \left({1}^{2} - 2 \cdot 1\right) = 1$

And the product of the roots

$= {\alpha}^{2015} {\beta}^{2015} = {\left(\alpha \beta\right)}^{2015} = {1}^{2015} = 1$

Nowthe quadratic equation whose roots are ${\alpha}^{2015} \mathmr{and} {\beta}^{2015}$ will be

x^2-("sum of the roots")x+"product of the roots"=0

$\implies {x}^{2} - x + 1 = 0$