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Given that $z = 3 + i$ is a root of $z^2 + pz + q = 0$, where $p$ and $q$ are real, what are the values of $p$ and $q$? - HSC - SSCE Mathematics Extension 2 - Question 2 - 2020 - Paper 1

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Given-that-$z-=-3-+-i$-is-a-root-of-$z^2-+-pz-+-q-=-0$,-where-$p$-and-$q$-are-real,-what-are-the-values-of-$p$-and-$q$?-HSC-SSCE Mathematics Extension 2-Question 2-2020-Paper 1.png

Given that $z = 3 + i$ is a root of $z^2 + pz + q = 0$, where $p$ and $q$ are real, what are the values of $p$ and $q$?

Worked Solution & Example Answer:Given that $z = 3 + i$ is a root of $z^2 + pz + q = 0$, where $p$ and $q$ are real, what are the values of $p$ and $q$? - HSC - SSCE Mathematics Extension 2 - Question 2 - 2020 - Paper 1

Step 1

Sub-part a: Identify the properties of the root

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Answer

Since z=3+iz = 3 + i is a complex root, its conjugate z∗=3−iz^* = 3 - i must also be a root of the equation. For the quadratic equation z2+pz+q=0z^2 + pz + q = 0, the sum and product of the roots can be used to find values for pp and qq.

Step 2

Sub-part b: Calculate the sum of the roots

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Answer

The sum of the roots is given by:

z+z∗=(3+i)+(3−i)=6z + z^* = (3+i) + (3-i) = 6

This implies:

ightarrow p = -6$$

Step 3

Sub-part c: Calculate the product of the roots

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Answer

The product of the roots is given by:

zimesz∗=(3+i)(3−i)=32−i2=9+1=10z imes z^* = (3+i)(3-i) = 3^2 - i^2 = 9 + 1 = 10

This implies:

q=10q = 10

Step 4

Final values of p and q

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Answer

Thus, the values are:

p=−6, q=10p = -6, \, q = 10

The correct answer is B.

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