Question

find the zeros of the quadratic polynomial 3x^2+9x+6 and verify the relationship between the zeros and the coefficient​

Answer 1

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Answer 2

$\underline\mathfrak{Given:-}$

• P (x) => 3x² + 9x + 6

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

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

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: P \: {(x)} \: \: = \: \: {3x}^{2} \: + \: {9x} \: + \: {6}$

$\: \: \: \: \: \leadsto \: \: {3x}^{2} \: + \: {9x} \: + \: {6}$

$\: \: \: \: \: \leadsto \: \: {3x}^{2} \: + \: {6x} \: + \: {3x} \: + \: {6}$

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

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

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

$\: \: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: {-1} \: \: \: and \: \: \: {x} \: \: = \: \: {-2}$

$\: \: \: \: \: \: \: \: \: {\alpha} \: \: = \: \: {-1} \: \: \: and \: \: \: {\beta} \: \: = \: \: {-2}$

$\underline\mathfrak{Verification:-}$

• P (x) => 3x² + 9x + 6
• a = 3
• b = 9
• c = 6

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

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

$\: \: \: \: \: \leadsto \: \: {-1} \: + \: {(-2)} \: \: = \: \: \cancel{\frac{-9}{3}}$

$\: \: \: \: \: \leadsto \: \: {-1} \: - \: {2} \: \: = \: \: {-3}$

$\: \: \: \: \: \leadsto \: \: {-3} \: \: = \: \: {-3}$

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

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

$\: \: \: \: \: \leadsto \: \: {-1} \: \times \: {(-2)} \: \: = \: \: \cancel{\frac{6}{3}}$

$\: \: \: \: \: \leadsto \: \: {2} \: \: = \: \: {2}$

Verified.