If 3 + 4i is a root of the equation {x^2} + p | vedclass

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(D) Given that the coefficients $p$ and $q$ are real numbers,the complex roots must occur in conjugate pairs.Since $3 + 4i$ is a root,its conjugate $3

Vedclass · Accepted Answer

$p = -6, q = 25$

Vedclass · Answer

(D) Given that the coefficients $p$ and $q$ are real numbers,the complex roots must occur in conjugate pairs.Since $3 + 4i$ is a root,its conjugate $3 - 4i$ must also be a root of the equation ${x^2} + px + q = 0$.For a quadratic equation ${x^2} + px + q = 0$,the sum of the roots is given by $-p$ and the product of the roots is given by $q$.Sum of roots: $(3 + 4i) + (3 - 4i) = 6 = -p \implies p = -6$.Product of roots: $(3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 - 16(-1) = 9 + 16 = 25 = q$.Thus,$p = -6$ and $q = 25$.

If $3 + 4i$ is a root of the equation ${x^2} + px + q = 0$ where $p$ and $q$ are real numbers,then:

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