Question

find the zeroes of the quadratic polynomial 6x2 - 13x + 6 and verify the relation between the zeroes and its coefficients​

Answer 1

Answer:

α = 3/2 and β =2/3

Step-by-step explanation:

6x² -13x + 6

6x² - (9+4)x + 6

6x² - 9x - 4x +6

3x(2x-3) -2(2x-3)

(3x-2)(2x-3)

3x=2 and 2x=3

x=2/3 and x=3/2

then, relation is

sum of zeroes = -b/a

2/3 + 3/2 = -(-13)/6

(4+9)/6 = 13/6

13/6 =13/6

product of zeroes = c/a

2/3 * 3/2 = 6/6

6/6 = 6/6

1 = 1

Answer 2

$\underline\mathfrak{Given:-}$

• 6x² - 13x + 6

$\underline\mathfrak{To \: \: Find:-}$

• Find the zeroes are coefficients ......?

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: P \: {(x)} \: \: = \: \: {6x}^{2} \: - \: {13} \: + \: {6}$

$\: \: \: \: \: \leadsto \: \: {6x}^{2} \: - \: {13} \: + \: {6}$

$\: \: \: \: \: \leadsto \: \: {6x}^{2} \: - \: {4x} \: - \: {9x} \: + \: {6}$

$\: \: \: \: \: \leadsto \: \: {2x} \: {({3x} \: - \: {2})} \: - \: {3} \: {({3x} \: - \: {2})}$

$\: \: \: \: \: \leadsto \: \: {({3x} \: - \: {2})} \: \: \: {({2x} \: - \: {3})}$

$\: \: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: \frac{2}{3} \: \: \: and \: \: \: {x} \: \: = \: \: \frac{3}{2}$

$\: \: \: \: \: \: \: \: \: {\alpha} \: \: = \: \: \frac{2}{3} \: \: \: and \: \: \: {\beta} \: \: = \: \: \frac{3}{2}$

$\underline\mathfrak{Verification:-}$

6x² - 13x + 6

• a = 3
• b = 8
• c = 6

$\: \: \: \: \: \therefore {Sum \: \: of \: \: zeroes} \: \: = \: \: \frac{ \: - \: coefficient \: \: of \: \: x}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: + \: {\beta} \: \: = \: \: \frac{-b}{a}$

$\: \: \: \: \: \leadsto \: \: \frac{2}{3} \: + \: \frac{3}{2} \: \: = \: \: - \: \frac{(-13)}{6}$

$\: \: \: \: \: \leadsto \: \: \frac{{4} \: + \: {9}}{6} \: \: = \: \: \frac{13}{6}$

$\: \: \: \: \: \leadsto \: \: \frac{13}{6} \: \: = \: \: \frac{13}{6}$

$\: \: \: \: \: \therefore {Product \: \: of \: \: zeroes} \: \: = \: \: \frac{constant \: \: term}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto \: \: {\alpha} \: {\beta} \: \: = \: \: \frac{c}{a}$

$\: \: \: \: \: \leadsto \: \: \frac{\cancel{2}}{\cancel{3}} \: \times \: \frac{\cancel{3}}{\cancel{2}} \: \: = \: \: \cancel{\frac{6}{6}}$

$\: \: \: \: \: \leadsto \: \: {1} \: \: = \: \: {1}$

Verified.