Question

1,$\alpha$,${\alpha }^2$,............................${\alpha }^6$ are the roots of the equation x = $(1{)}^{\frac{1}{7}}$.Find the quadratic equation whose roots are A = $\alpha$ + ${\alpha }^2$ + ${\alpha }^4$, B = ${\alpha }^3$ + ${\alpha }^5$ + ${\alpha }^6$.

• A. + x + 4 = 0
• B. + x + 2 = 0
• C. - x + 2 = 0
• D. - 2x + 2 = 0

Answer 1

The correct option is B

+ x + 2 = 0

Since 1,$\alpha$,${\alpha }^2$,${\alpha }^3$,${\alpha }^4$,${\alpha }^5$,${\alpha }^6$ are the roots of the equation

x = $(-1{)}^{\frac{1}{7}}$

1 + $\alpha$ + ${\alpha }^2$ + ${\alpha }^3$ + ${\alpha }^4$ + ${\alpha }^5$ + ${\alpha }^6$=0
$\alpha$ + ${\alpha }^2$ + ${\alpha }^3$ + ${\alpha }^4$ + ${\alpha }^5$ + ${\alpha }^6$ =-1

To form a quadratic equation

We needs to know sum of the roots and product of the roots of quadratic equation

So, let's find them

Roots of quadratic equation are A = $\alpha$ + ${\alpha }^2$ + ${\alpha }^4$ and B = ${\alpha }^3$ + ${\alpha }^5$ + ${\alpha }^6$

Sum of the root A + B = $\alpha$ + ${\alpha }^2$ + ${\alpha }^3$ + ${\alpha }^4$ + ${\alpha }^5$ + ${\alpha }^6$

A + B = -1 ----------(1)

Product of the roots of quadratic equation

A.B = ($\alpha$ + ${\alpha }^2$ + ${\alpha }^4$) (${\alpha }^3$ + ${\alpha }^5$ + ${\alpha }^6$) -------------(2)

A.B = ${\alpha }^4$ + ${\alpha }^6$ + ${\alpha }^7$ + ${\alpha }^5$ + ${\alpha }^7$ + ${\alpha }^8$ + ${\alpha }^7$ + ${\alpha }^9$ + ${\alpha }^{10}$

We know , ${\alpha }^2$ = 1

= ${\alpha }^4$ + ${\alpha }^6$ + 1 + ${\alpha }^5$ + 1 +$\alpha$ + 1 + ${\alpha }^2$ + ${\alpha }^3$

= ( 1 + $\alpha$ + ${\alpha }^2$ + ${\alpha }^3$ + ${\alpha }^4$ + ${\alpha }^5$ + ${\alpha }^6$) + 2

= 0 + 2

Therefore required equation is

$x^2$ - (A + B)x + A.B = 0

$x^2$ + x + 2 = 0