Question

.If p and q are zeroes of x2 + 5x + 8, find p+q
please help me​

Answer 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : x² + 5x + 8

Given that ,

⠀⠀⠀⠀⠀⠀⠀p and q are zeroes of Polynomial .

⠀⠀⠀⠀⠀⠀⠀Finding value of p + q :

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar\:\:\bf Sum \:of \: zeroes\:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} }\bigg\rgroup$

$\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}$

$\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ 5 \:)\:}{ 1 \:}$

$\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: - 5 \:\:}{ 1 \:}$

$\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \: - 5 \:\:$

$\qquad \dashrightarrow \underline{\pmb{\purple{\:p + q \: \: = \:\: \: - 5 \:\:\: }} }\:\bigstar$

⠀⠀⠀⠀⠀$\therefore {\underline{ \sf \:Hence,\:The\:value \:of\: p + q \:is\:\bf -5 }.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\qquad \qquad \boxed {\begin{array}{cc} \bf{\underline {\bigstar\:\: For \: a \:Quadratic \:Polynomial \::}}\\\\ \sf{ Whose \:\:zeroes \:\:are\:\:\alpha \:\&\;\: \beta\:\:} \\\\ 1)\:\: \alpha + \beta \: =\:\dfrac{-b}{a} \quad \bigg\lgroup \bf Sum\:of\;Zeroes \bigg\rgroup \\\\ 2)\:\: \alpha \times \beta \: =\:\dfrac{c}{a} \quad \bigg\lgroup \bf Product \:of\;Zeroes \bigg\rgroup \\\\ \end{array}}$

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Answer 2

$\huge\bf\fbox\red{Answer:-}$

⠀☆ GIVEN POLYNOMIAL : x² + 5x + 8

Given that ,

⠀⠀⠀⠀⠀⠀⠀p and q are zeroes of Polynomial .

⠀⠀⠀⠀⠀⠀⠀Finding value of p + q :

$\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar\:\:\bf Sum \:of \: zeroes\:\:: \end{gathered}$

$\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} }\bigg\rgroup \\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}\\ \\\end{gathered}$

⠀⠀⠀⠀⠀⠀$\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} \\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ 5 \:)\:}{ 1 \:} \\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: - 5 \:\:}{ 1 \:} \\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \: - 5 \:\: \\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \underline{\pmb{\purple{\:p + q \: \: = \:\: \: - 5 \:\:\: }} }\:\bigstar \\\\\end{gathered}$

⠀⠀⠀⠀⠀$\begin{gathered}\therefore {\underline{ \sf \:Hence,\:The\:value \:of\: p + q \:is\:\bf -5 }.}\\\\\end{gathered}$