Given equation can be rearranged as
x(x⁶ + 3x⁴ - 13x² - 15) = 0
Clearly x = 0 is one of the roots and the other part can be observed by replacing x² = t, from which we get:
t³ + 3t² - 13t - 15 = 0
⇒ (t - 3)(t² + 6t + 5) = 0
So, t = 3, t = -1, t = -5
Now we are getting:
x² = 3, x² = -1, x² = -5
⇒ x = ±√3, x = ±i, x = ±√5i
From the given condition:
|α₁| ≥ |α₂| ≥ .... ≥ |α₆|
We can clearly say that:
|α₁| = 0 and |α₂| = √5 = |α₅| and |α₄| = √3
= |α₃| and |α₂| = 1 = |α₁|
So we can have:
α₁ = √5i, α₂ = -√5i, α₃ = √3i, α₄ = √3, α₅ = i, α₆ = -i
Hence
α₁ - α₂ - α₃ + α₄ + α₅ + α₆ = 1 - (-3) + 5 = 9
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: