Question

The sum and product of the zeroes of a quadratic polynomial are 3 and -10 respectively the quadratic polynomial is

Answer 1

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

• ✦ sum of zeroes = 3
• ✦ product of zeroes = -10

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

• ✦ quadratic polynomial.

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

Let α and β be the zeroes of the required polynomial p(x).

Then

• (α + β) = 3
• αβ = -10

So,

p(x) = x² -(α + β)x + αβ

➝ p(x) = x² - 3x -10

So, the required polynomial is x² - 3x -10.

Verification:-

Sum of zeroes:-

(α + β) = - b/a

➝ 3 = -(-3)/1

➝ 3 = 3

Product of zeroes:-

αβ = c/a

➝ -10 = -10/1

➝ -10 = -10

LHS = RHS

hence, verified

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Answer 2

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

$\sf{Required \ polynomial \ is \ x^{2}+7x-30.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ sum \ and \ product \ of \ the}}$

$\sf{zeroes \ of \ a \ quadratic \ polynomial \ are}$

$\sf{3 \ and \ -10.}$

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

$\sf{Quadratic \ polynomial.}$

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

$\sf{Let \ \alpha \ be \ 3 \ and \ \beta \ be \ -10.}$

$\sf{Sum \ of \ zeroes=3+(-10)}$

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

$\sf{Product \ of \ zeroes=3(-10)}$

$\sf{\implies{\therefore{\alpha\beta=-30...(2)}}}$

$\sf{Quadratic \ polynomial \ is}$

$\sf{\implies{x^{2}-(\alpha+\beta)x+(\alpha\beta)}}$

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

$\sf{\implies{x^{2}+7x-30}}$

$\sf\purple{\tt{\therefore{Required \ polynomial \ is \ x^{2}+7x-30.}}}$

$\sf\blue{Verification:}$

$\sf{\implies{x^{2}+7x-30}}$

$\sf{\implies{x^{2}-3x+10x-30}}$

$\sf{\implies{x(x-3)+10(x-3)}}$

$\sf{\implies{(x-3)(x+10)}}$

$\sf{\implies{\therefore{x=3 \ or \ -10}}}$

$\sf{\therefore{Zeroes \ are \ 3 \ and \ -10}}$

$\sf{Hence, \ verified.}$