If 2-3i is a root of $z^{3} - 7 z^{2} + 25 z - 39 = 0$ $z^3-7z^2+25z-39=0$ , find the other two roots?

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If 2-3i is a root of $z^{3} - 7 z^{2} + 25 z - 39 = 0$ $z^3-7z^2+25z-39=0$ , find the other two roots?

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Answer 1 · 2017-06-06T10:04:18

2+3i, and 3.

Explanation:

I polynomial equations of order 3 (cubics like this) written in the form $a x^{3} + b x^{2} + c x + d = 0$ $ax^3 + bx^2 + cx + d = 0$ , with three roots, $α$ $alpha$ , $β$ $beta$ , and $γ$ $gamma$ , it is a rule that $α + β + γ = - b \div a$ $alpha + beta + gamma = -b-:a$ , also $α \cdot β + β \cdot γ + γ \cdot α = c \div a$ $alpha*beta + beta*gamma + gamma*alpha = c-:a$ , and that $α \cdot β \cdot γ = - d \div a$ $alpha*beta*gamma = -d-:a$ .
Now because $d$ $d$ is real and not imaginary, that means $α \cdot β \cdot γ$ $alpha*beta*gamma$ is real. You should know that to get a real number by multiplying imiginary numbers, you times it by it's conjugate, which for $2 - 3 i$ $2-3i$ is $2 + 3 i$ $2+3i$ . So now we know that $α$ $alpha$ and $β$ $beta$ are $2 - 3 i$ $2-3i$ and $2 + 3 i$ $2+3i$ .
We know $α \cdot β \cdot γ = - d \div a$ $alpha*beta*gamma = -d-: a$ , so $(2 - 3 i) (2 + 3 i) γ = - (- 39) \div 1$ $(2-3i)(2+3i)gamma = -(-39)-:1$ . Expanding $(2 - 3 i) (2 + 3 i)$ $(2-3i)(2+3i)$ we get $13$ $13$ , so $13 γ = 39$ $13gamma = 39$ , thus $γ = 39 \div 13 = 3$ $gamma = 39-:13 = 3$ , therefore giving us the other two roots: $2 + 3 i$ $2+3i$ and $3$ $3$ .

Answer 2 · 2017-06-06T12:57:20

$2 + 3 i and 3$ $2+3i" and " 3$

Explanation:

$the complex roots of polynomial equations always$ $"the complex roots of polynomial equations always "$
$occur in conjugate pairs$ $"occur in "color(blue)"conjugate pairs"$

$2 - 3 i is a root \Rightarrow 2 + 3 i is also a root$ $2-3i" is a root "rArr2+3i" is also a root"$

$the quadratic factor formed by these roots is$ $"the quadratic factor formed by these roots is "$

$(z - (2 - 3 i)) (z - (2 + 3 i))$ $(z-(2-3i))(z-(2+3i))$

$= ((z - 2) + 3 i) ((z - 2) - 3 i)$ $=((z-2)+3i)((z-2)-3i)$

$= {(z - 2)}^{2} - 9 i^{2}$ $=(z-2)^2-9i^2$

$= z^{2} - 4 z + 4 + 9$ $=z^2-4z+4+9$

$= z^{2} - 4 z + 13$ $=z^2-4z+13$

$\Rightarrow z^{3} - 7 z^{2} + 25 z - 39$ $rArrz^3-7z^2+25z-39$

$= z (z^{2} - 4 z + 13) + 4 z^{2} - 7 z^{2} - 13 z + 25 z - 39$ $=color(red)(z)(z^2-4z+13)color(magenta)(+4z^2)-7z^2color(magenta)(-13z)+25z-39$

$= z (z^{2} - 4 z + 13) - 3 (z^{2} - 4 z + 13) - 12 z + 12 z + 39$ $=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)color(magenta)(-12z)+12zcolor(magenta)(+39)$
$= - 39$ $color(white)(=)-39$

$= z (z^{2} - 4 z + 13) - 3 (z^{2} - 4 z + 13) + 0$ $=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)+0$

$\Rightarrow (z - 3) is a root$ $rArr(z-3)" is a root"$

$\Rightarrow (z - 3) (z - (2 - 3 i)) (z - (2 + 3 i)) = 0$ $rArr(z-3)(z-(2-3i))(z-(2+3i))=0$

$\Rightarrow roots are z = 3 and z = 2 \pm 3 i$ $rArr"roots are " z=3" and " z=2+-3i$

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If 2-3i is a root of $z^{3} - 7 z^{2} + 25 z - 39 = 0$ $z^3-7z^2+25z-39=0$ , find the other two roots?

2 Answers

Explanation:

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If 2-3i is a root of z^3-7z^2+25z-39=0z3−7z2+25z−39=0z^3-7z^2+25z-39=0, find the other two roots?

2 Answers

Explanation:

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If 2-3i is a root of $z^{3} - 7 z^{2} + 25 z - 39 = 0$ $z^3-7z^2+25z-39=0$ , find the other two roots?