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Sagot :
Answer:
1. (3z² - 27 = 0)
Rewriting in standard form:
[tex]3z^2 + 0z - 27 = 0 [/tex]
[tex]a = 3[/tex]
[tex]b = 0[/tex]
[tex]c = -27[/tex]
Sum of the roots:
[tex]\text{Sum} = -\frac{b}{a} = -\frac{0}{3} = 0 [/tex]
Product of the roots:
[tex]\text{Product} = \frac{c}{a} = \frac{-27}{3} = -9[/tex]
2. (z² + 8z + 12 = 0)
Already in standard form:
[tex]a = 1[/tex]
[tex]b = 8[/tex]
[tex]c = 12[/tex]
Sum of the roots:
[tex] \text{Sum} = -\frac{b}{a} = -\frac{8}{1} = -8[/tex]
Product of the roots:
[tex]\text{Product} = \frac{c}{a} = \frac{12}{1} = 12[/tex]
3. (6z² - 10z - 16 = 0)
Already in standard form:
[tex]a = 6[/tex]
[tex]b = -10[/tex]
[tex]c = -16[/tex]
Sum of the roots:
[tex]\text{Sum} = -\frac{b}{a} = -\frac{-10}{6} = \frac{10}{6} = \frac{5}{3}[/tex]
Product of the roots:
[tex]\text{Product} = \frac{c}{a} = \frac{-16}{6} = -\frac{8}{3}[/tex]
4. (12z² + 2 = 12z)
Rewriting in standard form:
[tex]12z^2 - 12z + 2 = 0[/tex]
[tex]a = 12[/tex]
[tex]b = -12[/tex]
[tex]c = 2[/tex]
Sum of the roots:
[tex]\text{Sum} = -\frac{b}{a} = -\frac{-12}{12} = 1[/tex]
Product of the roots:
[tex]\text{Product} = \frac{c}{a} = \frac{2}{12} = \frac{1}{6}[/tex]
5. (z(2z² + 3) = 3z + 1)
First, simplify the equation:
[tex]2z^3 + 3z = 3z + 1[/tex]
[tex]2z^3 + 3z - 3z - 1 = 0[/tex]
[tex]2z^3 - 1 = 0[/tex]
This simplifies to:
[tex]2z^3 = 1[/tex]
Since this is not a quadratic equation, we can't use the formulas for sum and product of the roots directly.
However, for the sake of completeness, considering only expressions of the form (az² + bz + c = 0):
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