Question

which is
the sum of zeros of the quadratic polynomial 6x²-4x+3​

Answer 1

Answer:

$= 6x^{2} - 4x + 3$

a = 6

b = -4

c = 3

$\frac{-b ± \sqrt{b^{2} - 4ac}}{2a}\\\\= \frac{-(-4) ± \sqrt{(-4)^{2} - 4(6)(3)}}{2(6)}\\\\= \frac{4 ± \sqrt{16 - 72}}{12}\\\\= \frac{4 ± \sqrt{-56}}{12}\\\\= \frac{4 ± (-2\sqrt{14})}{12}\\\\= \frac{2(2 ± (-\sqrt{14}))}{12}\\\\= \frac{2 ± (-\sqrt{14})}{6}$

thus

$x = \frac{2 + \sqrt{14}}{6}$            &                 $x = \frac{2 - \sqrt{14}}{6}$

therefore the sum of the zeros are

$= (\frac{2 + \sqrt{14}}{6} )+(\frac{2 - \sqrt{14}}{6})\\\\= \frac{2 +\sqrt{14} + 2 - \sqrt{14}}{6}\\\\=\frac{4}{6}\\\\= \frac{2}{3}$

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Answer 2

Answer:

I don't know this answer ask someone else