If x^2-x+1=0, the value of ∑n=1^5(x^n+(1)/(x^n))^2 is (a) 8 (b) 10 (c) 12 (d) 5

Name: Problem 47, Chapter 1: IIT JEE Super Course in Mathematics ALGEBRA II
Uploaded: 2021-06-05T09:37:36-08:00
Duration: 10 min 8 s
Channel: Anas Venkitta
Description: If x^2-x+1=0, the value of ∑n=1^5(x^n+(1)/(x^n))^2 is (a) 8 (b) 10 (c) 12 (d) 5

If x^2-x+1=0, the value of ∑n=1^5(x^n+(1)/(x^n))^2 is (a) 8...

Key Concepts

De Moivre's Theorem

De Moivre's theorem connects complex numbers expressed in polar form with trigonometric functions. It states that (cos? + i sin?)? = cos(n?) + i sin(n?), making it a powerful tool for calculating powers of complex numbers and simplifying expressions involving sums of complex exponentials.

Vieta's Formulas

Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. In the context of quadratic equations, these formulas provide a quick way to compute expressions like the sum and product of the roots without explicitly solving for them.

Quadratic Equations

A quadratic equation is a second?degree polynomial equation in a single variable. Its standard form is ax² + bx + c = 0 and its solutions can be found using the quadratic formula. The nature of the solutions (real or complex) depends on the discriminant, b² - 4ac.

Complex Numbers

Complex numbers extend the real numbers with the addition of the imaginary unit i, where i² = -1. They arise naturally as solutions to quadratic equations with negative discriminants and have both a rectangular form (a + bi) and a polar form, allowing for easier manipulation, especially when raising to powers.

Transcript

00:01 So in the given question, we are told that if x squared, if x squared minus x plus 1 is equal to 0, then we are told to find the value of the expression sigma n equal to 1 to 5 of x raised to the power n plus 1 by x raised to the power n squared.

00:30 Right so this is the expression that we have to evaluate so what we should know first is there is omega which is the complex cube root of one and one of the properties of this omega is that one plus omega plus omega squared gives as the result as zero so we have all we already know this relation so, we have to compare this relation with the given expression, right? so, when we compare, we can see that when we substitute the value of x over here as minus omega, what we get is omega square plus omega plus one, right? so we get this relation when we substitute x equal to omega in the above result, above expression.

01:30 So we can say that we can take the value of x as minus omega since it is the solution of the given equation.

01:38 So now that we have the value of x, we can simplify the expression that we need to find the value of.

01:45 So we have sigma n equal to 1 to 5 of x square x x x raised to n square plus two times x raised to n times 1 by x raised to n plus 1 by x raised to n whole square right so this is what we get after expanding this and we can simplify this further so we will have sigma n equal to 1 to 5 x raised to the power 2 n plus x n x x x raised 2 n and x raised 2n and x w cc cancels off from the second term as they are in the numerator and denominator so plus 2 plus 1 by x raised to 2 right this is what we have and now let's substitute the value of x as omega right x as minus omega in this result so what we have is sigma n equal to 1 to 5 omega square omega raised to 2 n plus 2 plus 1 by omega raised 2 in right and what we are going to do next is we should know the property of omega that 1 by omega is equal to omega square so using this property we can write 1 by omega raised to the power 2n as we can write 1 by omega raised to the power 2n as 1 by omega whole raised to the power 2n and we can substitute 1 by omega as omega square which means we will have omega squared raised to the power 2n which is equal to omega raised to 4m right so we are using this substitution over here that 1 by omega raised to 2n is equal to omega raised to the power 4n...

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