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Find the zeros: 3r^2 − 16r − 12 = 0

Sagot :

djtwinx017

Answer:

r = 6, r = - ⅔

Step-by-step explanation:

Given the quadratic equation, 3r² - 16r - 12 = 0:

where a = 3, b = -16, and c = -12.

Use the quadratic formula to find the x-intercepts:

[tex]r = \frac{-b +/- \sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]r = \frac{16 +/- \sqrt{(-16)^{2}-4(3)(-12) } }{2(3)}[/tex]

[tex]r = \frac{16 +/- \sqrt{256 - 144 } }{6}[/tex]

[tex]r = \frac{16 +/- \sqrt{400} }{6}[/tex]

[tex]r = \frac{16 + 20}{6} = \frac{36}{6} = 6[/tex] ,  [tex]r = \frac{16 - 20}{6} = - \frac{4}{6} = -\frac{2}{3}[/tex]

Therefore, the roots (zeroes) of the given quadratic equation are:

r = 6, r = - ⅔

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