The equation 6x^4-5x^3+13x^2-5x+6=0 will have | vedclass

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(B) Given the reciprocal equation $6x^4-5x^3+13x^2-5x+6=0$. Dividing by $x^2$ (since $x=0$ is not a root),we get $6x^2-5x+13-\frac{5}{x}+\frac{6}{x^2}

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only complex roots

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(B) Given the reciprocal equation $6x^4-5x^3+13x^2-5x+6=0$. Dividing by $x^2$ (since $x=0$ is not a root),we get $6x^2-5x+13-\frac{5}{x}+\frac{6}{x^2}=0$. Grouping terms: $6(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+13=0$. Let $t = x+\frac{1}{x}$,then $x^2+\frac{1}{x^2} = t^2-2$. Substituting: $6(t^2-2)-5t+13=0 \implies 6t^2-5t+1=0$. Solving for $t$: $(3t-1)(2t-1)=0$,so $t=\frac{1}{3}$ or $t=\frac{1}{2}$. For $x+\frac{1}{x} = \frac{1}{3}$,$x^2-\frac{1}{3}x+1=0$. The discriminant $D = (\frac{1}{3})^2 - 4(1) = \frac{1}{9}-4 For $x+\frac{1}{x} = \frac{1}{2}$,$x^2-\frac{1}{2}x+1=0$. The discriminant $D = (\frac{1}{2})^2 - 4(1) = \frac{1}{4}-4 Since both quadratic equations have negative discriminants,all four roots are complex.

The equation $6x^4-5x^3+13x^2-5x+6=0$ will have

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