The number of non-real roots of the equation | vedclass

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Question

(A) Given equation is $x^{10}-3x^8+5x^6-5x^4+3x^2-1=0$. Let $f(x) = x^{10}-3x^8+5x^6-5x^4+3x^2-1$. We can factorize the expression by grouping terms:

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Given equation is $x^{10}-3x^8+5x^6-5x^4+3x^2-1=0$. Let $f(x) = x^{10}-3x^8+5x^6-5x^4+3x^2-1$. We can factorize the expression by grouping terms: $f(x) = (x^{10}-1) - 3x^2(x^6-1) + 5x^4(x^2-1)$. Using the identity $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + b^{n-1})$,we can see that $(x^2-1)$ is a common factor. $f(x) = (x^2-1)(x^8+x^6+x^4+x^2+1) - 3x^2(x^2-1)(x^4+x^2+1) + 5x^4(x^2-1)$. $f(x) = (x^2-1)[(x^8+x^6+x^4+x^2+1) - 3x^2(x^4+x^2+1) + 5x^4]$. $f(x) = (x^2-1)[x^8+x^6+x^4+x^2+1 - 3x^6-3x^4-3x^2 + 5x^4]$. $f(x) = (x^2-1)(x^8-2x^6+3x^4-2x^2+1)$. $f(x) = (x^2-1)(x^4-x^2+1)^2$. Setting $f(x) = 0$,we get $x^2-1=0$ or $(x^4-x^2+1)^2=0$. $x^2-1=0 \implies x = \pm 1$ ($2$ real roots). For $x^4-x^2+1=0$,let $t = x^2$. Then $t^2-t+1=0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 Thus,$t = \frac{1 \pm i\sqrt{3}}{2}$. Since $x^2 = t$,each value of $t$ gives $2$ complex roots for $x$. Since there are $2$ values of $t$,we get $2 	imes 2 = 4$ complex roots from the squared term. However,the term $(x^4-x^2+1)^2$ implies each root is repeated twice,so there are $8$ non-real roots in total.

The number of non-real roots of the equation $x^{10}-3x^8+5x^6-5x^4+3x^2-1=0$ is

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