Question

Find the zeroes of the quadratic polynomial f(x) = x2–3x –28 and verify the relationship between the zeroes and the coefficients of the polynomial.

Answer 1

Answer:

$Solution : </p><p></p><p>x2−3x−10=0x2-3x-10=0 </p><p>x2−5x+2x−10=0x2-5x+2x-10=0 </p><p>x(x−5)+2(x−5)=0x(x-5)+2(x-5)=0 </p><p>(x−5)(x+2)=0(x-5)(x+2)=0 </p><p>X=5,−2X=5,-2 </p><p>Sum of the roots =−ba=31=-ba=31 </p><p>which is same as 5−2=35-2=3 </p><p>product of the roots =ca=−10=ca=-10 </p><p>which is same as 5x(−2)=−105x(-2)=-10 </p><p>Hence verified</p><p></p><p>$

Answer 2

GIVEN⠀ POLYNOMIAL  – x² - 3x  - 28 :

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$:\implies\sf x^2 -3x - 28 = 0 \\\\\\:\implies\sf x^2 + 4x - 7x -28 = 0 \\\\\\:\implies\sf x(x + 4) -7(x + 4) = 0\\\\\\:\implies\sf (x - 7)\; (x + 4) = 0\\\\\\:\implies{\underline{\boxed{\frak{\purple{x = 7\; or \; -4}}}}}\;\bigstar$

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❍$\underline{\textsf{Relation b/w Coefficients \& Zeroes :}}$

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${\qquad\maltese\:\:\textsf{Sum of Zeroes :}} \\\\\dashrightarrow\sf\:\:\alpha +\beta= \dfrac{ - \:b}{a}\\\\\\\dashrightarrow\sf\:\:7+(-4) =\dfrac{-(-3)}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf\:3=\:3}}\\\\\\{\qquad\maltese\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{c}{a}\\\\\\\dashrightarrow\sf\:\:(7)\:(-4)=\dfrac{-28}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf -\:28=-\:28}}$

$\rule{300}{1.5}$

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$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Quadratic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^\sf2 \sf + bx + c}\\\sf with zeroes \alpha\:\sf and\:\beta \\\\\\ {\textcircled{\footnotesize1}} \:\:\alpha +\beta= \dfrac{ - \:b}{a}\:\:\bigg\lgroup\bf Sum\:of\:Zeroes\bigg\rgroup \\\\\\{\textcircled{\footnotesize2}} \: \:\alpha \beta= \sf\dfrac{c}{a}\:\:\bigg\lgroup\bf Product\:of\:Zeroes\bigg\rgroup\end{minipage}}$

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NOTE: See the answer from web. :) ⠀

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