If (2+i) is a root of the equation x^3-5x^2+9 | vedclass

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(A) Given that $(2+i)$ is a root of the polynomial equation $x^3-5x^2+9x-5=0$ with real coefficients,its complex conjugate $(2-i)$ must also be a root

Vedclass · Accepted Answer

$1$ and $(2-i)$

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(A) Given that $(2+i)$ is a root of the polynomial equation $x^3-5x^2+9x-5=0$ with real coefficients,its complex conjugate $(2-i)$ must also be a root.Let the three roots be $r_1 = 2+i$,$r_2 = 2-i$,and $r_3 = \alpha$.According to the properties of roots of a cubic equation $ax^3+bx^2+cx+d=0$,the product of the roots is given by $r_1 r_2 r_3 = -d/a$.Here,$a=1$ and $d=-5$,so $r_1 r_2 r_3 = -(-5)/1 = 5$.Substituting the known roots: $(2+i)(2-i) \alpha = 5$.Since $(2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5$,we have $5 \alpha = 5$,which implies $\alpha = 1$.Thus,the other roots are $1$ and $(2-i)$.

If $(2+i)$ is a root of the equation $x^3-5x^2+9x-5=0$,then the other roots are

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