Math, asked by kavithachilukuri1, 4 months ago

find the zeroes of the quadratic polynomial x^2+5x+6 and verify the relationship between the zeroes and Coffecients​

Answers

Answered by ShírIey
116

GIVEN POLYNOMIAL x² + 5 + 6 :

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:\implies\sf x^2 + 5x + 6 = 0 \\\\\\:\implies\sf x^2 + 2x + 3x + 6 = 0 \\\\\\:\implies\sf  x(x + 2) +3(x + 2) = 0\\\\\\:\implies\sf (x + 3)\; (x + 2) = 0\\\\\\:\implies{\underline{\boxed{\frak{\purple{x = -2 \; or \; -3}}}}}\;\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\:\:-2-3 =\dfrac{-5}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf-\:5=-\:5}}\\\\\\{\qquad\maltese\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{c}{a}\\\\\\\dashrightarrow\sf\:\:(-2)\:(-3)=\dfrac{6}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf 6=\:6}}

\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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Answered by Anonymous
90

Answer:

Given :-

  • x² + 5x + 6

To Find :-

  • What are the zeros of the quadratic polynomial and verify the relationship between the zeros and coffecients.

Solution :-

First, we have to find the value of x :

\longmapsto \sf {x}^{2} + 5x + 6

\sf {x}^{2} + (3 + 2)x + 6 =\: 0

\sf {x}^{2} + 3x + 2x + 6 =\: 0

\sf x(x + 3) + 2(x + 3) =\: 0

\sf x + 3 =\: 0

\sf\bold{\green{x =\: -\: 3}}

Either,

\sf x + 2 =\: 0

\sf\bold{\green{x =\: -\: 2}}

Then we get,

\implies \sf \alpha =\: -\: 2 , \beta =\: -\: 3

Now, we have to find the relationship between the zeros and coffecients :

Given polynomial :

\leadsto \sf {x}^{2} + 5x + 6

where,

  • a = 1
  • b = 5
  • c = 6

To find sum of zeros we know that,

\sf\boxed{\bold{\alpha + \beta =\: \dfrac{- b}{a}}}

\sf \alpha + \beta =\: \dfrac{- 5}{1}

\sf (- 2) + (- 3) =\: - 5

\sf - 2 - 3 =\: - 5

\sf\bold{\red{- 5 =\: - 5}}

Again, to find product of zeros we know that,

\sf\boxed{\bold{\alpha\beta =\: \dfrac{c}{a}}}

\sf \alpha\beta =\: \dfrac{6}{1}

\sf (- 2)(- 3) =\: 6

\sf - 2 \times - 3 =\: 6

\sf\bold{\red{6 =\: 6}}

\longrightarrow {\red{\bigstar}} \ {\underline{\green{\textsf{\textbf{HENCE, VERIFIED}}}}}

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