Question

Find the zeroes of quadratic polynomial and verify the relation between the zeros and its coefficients 1. X Sq-1 by 6 x -2​

Answer 1

Step-by-step explanation:

${x}^{2} - \frac{1}{6} x - 2 = 0 \\ \\ \frac{6 {x}^{2} - x - 12}{6} = 0 \\ \\ 6 {x}^{2} - x - 12 = 0 \\ \\ 6x {}^{2} - 9x + 8x - 12 = 0 \\ \\ 3x(2x - 3) + 4(2x - 3) = 0 \\ \\ (2x - 3)(3x + 4) = 0$

Therefore,

$x = \dfrac{3}{2} \: \: or \: \: x = \dfrac{ - 4}{3}$

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6x² - x - 12, here

• a = 6
• b = - 1
• c = - 12

Verification ;

$\mathrm{sum \: of \: zeroes = \frac{ - b}{a} } \\ \\ \implies \frac{3}{2} - \frac{4}{3} = \frac{1}{6} \\ \\ \implies \frac{9 - 8}{6} = \frac{1}{6} \\ \\ \implies \frac{1}{6} = \frac{1}{6}$

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$\mathrm{product \: of \: zeroes = \frac{c}{a} } \\ \\ \implies \frac{3}{2} \times ( \frac{ - 4}{3} ) = \frac{ - 12}{6} \\ \\ \implies - 2 = - 2$

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Thanks for the question!!

Answer 2

Step-by-step explanation:

$= \frac{ {x}^{2} - 1}{6x - 2}$

The constant term in this polynomial is 2

$\frac{ ({ - 1})^{2} - 1 }{ - 6 - 2} \\ = \frac{1 - 1}{ - 4} \\ = \frac {0}{ - 4} \\ = 0$

(as it has no variable)

The factors of 2 are 1 and 2

Considering the first factor, which is 1,

Let X + 1= 0

X= (-1)

Substitute the value in the polynomial

Hence, p(X) = 0

Thus, we conclude that the zero of this polynomial is -1

It's coefficients are 1 and 6