Question

find the zeroes of the quadric polynomial 6x2+x-2 and verify the relationship between the zeroes and the cofficiant​

Answer 1

Answer:

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6x2 - 7x - 3 = 0 6x2 + 2x - 9x - 3 = 0 2x (3x + 1) - 3 (3x + 1) = 0 (2x - 3)(3x + 1) = 0 32, -13 are zeroes Sum of zeroes = 32 - 13 = 9 - 26 = 76 ...

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Answer 2

Answer:

$6 {x}^{2} + x - 2 = 0 \\ \\ 6 {x}^{2} - 3x + 4x - 2 = 0 \\ \\ 3x(2x - 1) + 2(2x - 1) = 0 \\ \\ (2x - 1) \: (3x + 2) = 0$

When 2x - 1 = 0

$2x = 1 \\ \\ x = \frac{1}{2} \: \: (say \: \alpha )$

When 3x + 2 = 0

$3x = - 2 \\ \\ x = \frac{ - 2}{3} \: \: (say \: \beta )$

Now

$sum \: of \: zeroes \: ( \alpha + = \beta ) = \frac{1}{2} - \frac{2}{3} \\ \\ = \frac{3 - 4}{6} \\ \\ = \frac{ - 1}{6} = \frac{ - b}{a}$

$product \: of \: zeroes( \alpha \beta ) = \frac{1}{2} \times \frac{ - 2}{3} \\ \\ = \frac{ -1 }{3} = \frac{c}{a}$

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