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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

Substituting the Root

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Answer

To find the values of pp and qq, substitute z=3+iz = 3 + i into the equation z2+pz+q=0z^2 + pz + q = 0.

Step 2

Calculating $z^2$

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Answer

First, compute z2z^2: $$z^2 = (3 + i)^2 = 3^2 + 2(3)(i) + i^2 = 9 + 6i - 1 = 8 + 6i.$$$$

Step 3

Setting Up the Equation

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Answer

Now substitute this back into the equation: 8+6i+p(3+i)+q=0.8 + 6i + p(3 + i) + q = 0.

Step 4

Separating Real and Imaginary Parts

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Answer

This gives us: (8+3p+q)+(6+p)i=0.(8 + 3p + q) + (6 + p)i = 0. For the equation to hold, both the real and imaginary parts must equal zero:

  1. 8+3p+q=08 + 3p + q = 0 (Real Part)
  2. 6+p=06 + p = 0 (Imaginary Part)

Step 5

Solving for $p$

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Answer

From the imaginary part equation 6+p=06 + p = 0, we find: p=−6.p = -6.

Step 6

Substituting $p$ to Find $q$

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Answer

Now substitute p=−6p = -6 into the real part equation: 8+3(−6)+q=08 + 3(-6) + q = 0 This simplifies to:

ightarrow q = 10.$$

Step 7

Final Values

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114 rated

Answer

Thus, the values are: p=−6,q=10.p = -6, q = 10. This matches with option B.

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