Question

find the quadratic polynomial whose zeroes are -2 and -5. verify the relationship between zeroes and cofficient of polynomial​

Answer 1

$\large{\underline{\bf{\purple{Given:-}}}}$

• given zeroes of required polynomial are -2 and -5

$\large{\underline{\bf{\purple{To\:Find:-}}}}$

• we need to find the polynomial and also find the relationship between the zeroes and coefficients.

$\huge{\underline{\bf{\red{Solution:-}}}}$

• Let α be -2 and β be -5

α + β = -2+(-5)

⠀⠀⠀➝ -7

αβ = -2 × -5

⠀⠀⠀➝ 10

• p(x) = x²-(α + β)x +αβ

⠀⠀⠀➝ x²-(-7)x +10

⠀⠀⠀➝ x² + 7x + 10

So, the required polynomial is x² + 7x + 10.

Relationship between the zeroes and coefficients:-

Sum of zeroes = coefficient of x /coefficient ofx²

• (α + β) = - b/a

⠀⠀⠀➝ -7 = -7/1

⠀⠀⠀➝ -7 = -7

Product of zeroes = constant term/coefficient of x²

• αβ = c/a

⠀⠀⠀➝ 10 = 10/1

⠀⠀⠀➝ 10 = 10

LHS = RHS

Hence relationship is verified

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Answer 2

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ required \ polynomial \ is \ x^{2}+7x+10.}$

$\sf\orange{Given:}$

$\sf{\implies{Zeroes \ of \ polynomial \ are \ -2 \ and \ -5}}$

$\sf\pink{To \ find:}$

$\sf{Quadratic \ polynomial \ with \ zeroes \ -2 \ and}$

$\sf{-5.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ \alpha \ be \ -2 \ and \ beta \ be \ -5}$

$\sf{Sum \ of \ zeroes=-2-5}$

$\sf{\implies{\therefore{\alpha+\beta=-7...(1)}}}$

$\sf{Product \ of \ zeroes=-2\times(-5)}$

$\sf{\implies{\therefore{\alpha\beta=10...(2)}}}$

$\sf{Quadratic \ polynomial \ is}$

$\sf{x^{2}-(Sum \ of \ zeroes)x+(Product \ of \ zeroes)}$

$\sf{...from \ (1) \ and \ (2)}$

$\sf{\implies{x^{2}+7x+10}}$

$\sf\purple{\tt{\therefore{The \ required \ polynomial \ is \ x^{2}+7x+10.}}}$

$\bold\blue{\underline{\underline{Solution:}}}$

$\sf{\implies{x^{2}+7x+10}}$

$\sf{Here, \ a=1, \ b=7 \ and \ c=10}$

$\sf{\frac{-b}{a}=-7}$

$\sf{Also, \ \alpha+\beta=-7}$

$\sf{\therefore{Sum \ of \ zeroes=\frac{-b}{a}}}$

$\sf{\frac{c}{a}=10}$

$\sf{Also, \ \alpha\beta=10}$

$\sf{\therefore{Product \ of \ zeroes=\frac{c}{a}}}$

$\sf{Hence, \ verified. }$