Question

find the zeros of the following quadratic polynomials by factorization method and verify the relation between the zeros and tne coefficients: 6x^2 + x - 12

Answer 1

Let 6x2 - 7x-3 be p(x) =0

so,6x2 - 7x-3 =0

6x2 - (9-2) x -3

6x2 - 9x + 2x -3

(6x2 - 9x)(2x-3)

3x(2x-3) +1 (2x-3)

(3x +1) (2x -3) = 0

⇒ 3x +1 = 0

⇒ 2x -3 = 0

⇒ x =3/2

⇒ x = - 1/3

sum of zeroes = -1/3 + 3/2 =7/6 = -b/a

product of zeroes =- 1/3 * 3/2 = - 3/6 or -1/2 = c/a

Answer 2

Answer:

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

$\alpha = \frac{ (- 3)}{2} \\ \beta = \frac{4}{3}$

${ax}^{2} + bx + c = {6x}^{2} + x - 12 \\ a = 6 \\ b = 1 \\ c =( - 12)$

$\alpha + \beta = \frac{( - b)}{a} \\ \alpha \beta = \frac{c}{a}$

$\alpha + \beta = \frac{ (- b)}{a} \\ \frac{( - 3)}{2} + \frac{4}{3} = \frac{ (- 1)}{6} \\ \frac{( - 3 \times 3 ) + (4 \times 2)}{6} = \frac{( - 1)}{6} \\ \frac{( - 9) + (8)}{6} = \frac{ - 1)}{6} \\ \frac{( - 1)}{6} = \frac{( - 1)}{6}$

$( \alpha \beta = \frac{c}{a} \\ \frac{( - 3)}{2} \times \frac{4}{3} = \frac{( - 12)}{6} \\ \frac{( - 12)}{6} = \frac{( - 12)}{6} \\ ( - 2) = ( - 2)$

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