Question

Find the zeroes of the quadratic polynomial x2 + 9x + 18, and verify the relationship between the zeroes and the coefficients.

Answer 1

x^2+9x+18

By factorization

x^2+3x+6x+18

x(x+3)+6(x+3)

x+6=0. x+3=0

x=-6. x=-3

alpha=-6. Beta=-3

Relationship between zeroes and coefficients

Sum of zeroes(alpha+beta)=-6+(-3)=-6-3=-9

Product of zeroes(alpha×beta)=18×1=18

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Answer 2

$\underline\mathfrak{Given:-}$

• P (x) => x² + 9x + 18

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

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

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: P \: {(x)} \: \: = \: \: {x}^{2} \: + \: {9x} \: + \: {18}$

$\: \: \: \: \: \leadsto \: \: {x}^{2} \: + \: {9x} \: + \: {18}$

$\: \: \: \: \: \leadsto \: \: {x}^{2} \: + \: {6x} \: + \: {3x} \: + \: {18}$

$\: \: \: \: \: \leadsto \: \: {x} \: {({x} \: + \: {6})} \: + \: {3} \: {({x} \: + \: {6})}$

$\: \: \: \: \: \leadsto \: \: {({x} \: + \: {3})} \: \: \: {({x} \: + \: {6})}$

$\: \: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: {-3} \: \: \: and \: \: \: {x} \: \: = \: \: {-6}$

$\: \: \: \: \: \: \: \: \: {\alpha} \: \: = \: \: {-3} \: \: \: and \: \: \: {\beta} \: \: = \: \: {-6}$

$\underline\mathfrak{Verification:-}$

• x² + 9x + 18
• a = 1
• b = 9
• c = 18

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

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

$\: \: \: \: \: \leadsto \: \: {-3} \: + \: {(-6)} \: \: = \: \: {-9}$

$\: \: \: \: \: \leadsto \: \: {-3} \: - \: {6} \: \: = \: \: {-9}$

$\: \: \: \: \: \leadsto \: \: {-9} \: \: = \: \: {-9}$

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

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

$\: \: \: \: \: \leadsto \: \: {-3} \: \times \: {(-6)} \: \: = \: \: {18}$

$\: \: \: \: \: \leadsto \: \: {18} \: \: = \: \: {18}$

Verified.