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Determine the sum and product of the roots of each quadratic equation. show your solution

1. 3z² - 27 = 0?

2. z² + 8z +1 2 = 0?

3. 6z² - 10z - 16 = 0?

4. 12z² + 2 = 12z?

5. z(2z² + 3) = 3z + 1?


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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