Let $α, β, γ, δ$ be the roots of the equation $x^{4} - x^{3} - x^{2} - 1 = 0.$ Also consider $p (x) = x^{6} - x^{5} - x^{3} - x^{2} - x,$ then the value of $p (α) + p (β) + p (γ) + p (δ)$ cannot be:

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $4$
B $5$
D $7$
$p (x) = x^{6} - x^{5} - x^{3} - x^{2} - x = x^{6} - x^{5} - x^{3} - x^{2} - x + x^{4} - x^{4} = x^{2} (x^{4} - x^{3} - x^{2}) + (x^{4} - x^{3} - x^{2}) - x = x^{2} - x + 1$

$p (α) + p (β) + p (γ) + p (δ) = \sum α^{2} - \sum α + 4 = (α + β + γ + δ)^{2} - 2 (α β + β γ + γ δ + α δ + α γ + β δ) - \sum α + 4 = 6$

Suggest Corrections

Similar questions

Fill all the roots of the equation.

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

Given equation $x^{6}$ - $x^{5}$ + $x^{4}$ - $x^{3}$ + $x^{2}$ - x + 1 = 0 can also be written as _________.

where 1 + x ≠ 0

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}

α

x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6} < x_{7} < x_{8}

x_{3}, x_{4}, x_{5}, x_{6}

x_{1}, x_{2}, x_{3}, x_{4}

x_{5}

p (x) = x^{5} + x^{2} + 1

g (x) = x^{2} - 2

g (x_{1}) g (x_{2}) g (x_{3}) g (x_{4}) g (x_{5}) - 30 g (x_{1} x_{2} x_{3} x_{4} x_{5})

Let f (x) = x^{6} - 2 x^{5} + x^{3} + x^{2} - x - 1

g (x) = x^{4} - x^{3} - x^{2} - 1

Let a, b, c and d the roots of g (x) = 0

f (a) + f (b) + f (c) + f (d)

Join BYJU'S Learning Program