⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : x² + 5x + 8 Given that , ⠀⠀⠀⠀⠀⠀⠀p and q are zeroes of Polynomial . ⠀⠀⠀⠀⠀⠀⠀Finding value of p + q : [tex]\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar\:\:\bf Sum \:of \: zeroes\:\:: [/tex] [tex]\qquad \dag\:\:\bigg\lgroup \sf{ p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} }\bigg\rgroup \\\\[/tex] [tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}\\ \\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} \\\\[/tex] [tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ 5 \:)\:}{ 1 \:} \\\\[/tex] [tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: – 5 \:\:}{ 1 \:} \\\\[/tex] [tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \: – 5 \:\: \\\\[/tex] [tex]\qquad \dashrightarrow \underline{\pmb{\purple{\:p + q \: \: = \:\: \: – 5 \:\:\: }} }\:\bigstar \\\\[/tex] ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \sf \:Hence,\:The\:value \:of\: p + q \:is\:\bf -5 }.}\\\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ⠀⠀⠀⠀⠀[tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex] [tex]\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}} [/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
[tex]\huge\bf\fbox\red{Answer:-}[/tex] ⠀☆ GIVEN POLYNOMIAL : x² + 5x + 8 Given that , ⠀⠀⠀⠀⠀⠀⠀p and q are zeroes of Polynomial . ⠀⠀⠀⠀⠀⠀⠀Finding value of p + q : [tex]\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar\:\:\bf Sum \:of \: zeroes\:\:: \end{gathered}[/tex] [tex]\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} }\bigg\rgroup \\\\\end{gathered}[/tex] [tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}\\ \\\end{gathered}[/tex] ⠀⠀⠀⠀⠀⠀[tex]\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\end{gathered}[/tex] [tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} \\\\\end{gathered}[/tex] [tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ 5 \:)\:}{ 1 \:} \\\\\end{gathered} [/tex] [tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: – 5 \:\:}{ 1 \:} \\\\\end{gathered}[/tex] [tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \: – 5 \:\: \\\\\end{gathered}[/tex] [tex]\begin{gathered}\qquad \dashrightarrow \underline{\pmb{\purple{\:p + q \: \: = \:\: \: – 5 \:\:\: }} }\:\bigstar \\\\\end{gathered}[/tex] ⠀⠀⠀⠀⠀[tex]\begin{gathered}\therefore {\underline{ \sf \:Hence,\:The\:value \:of\: p + q \:is\:\bf -5 }.}\\\\\end{gathered}[/tex] Reply
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : x² + 5x + 8
Given that ,
⠀⠀⠀⠀⠀⠀⠀p and q are zeroes of Polynomial .
⠀⠀⠀⠀⠀⠀⠀Finding value of p + q :
[tex]\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar\:\:\bf Sum \:of \: zeroes\:\:: [/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} }\bigg\rgroup \\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}\\ \\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} \\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ 5 \:)\:}{ 1 \:} \\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: – 5 \:\:}{ 1 \:} \\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \: – 5 \:\: \\\\[/tex]
[tex]\qquad \dashrightarrow \underline{\pmb{\purple{\:p + q \: \: = \:\: \: – 5 \:\:\: }} }\:\bigstar \\\\[/tex]
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \sf \:Hence,\:The\:value \:of\: p + q \:is\:\bf -5 }.}\\\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀⠀⠀[tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex]
[tex]\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}} [/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
[tex]\huge\bf\fbox\red{Answer:-}[/tex]
⠀☆ GIVEN POLYNOMIAL : x² + 5x + 8
Given that ,
⠀⠀⠀⠀⠀⠀⠀p and q are zeroes of Polynomial .
⠀⠀⠀⠀⠀⠀⠀Finding value of p + q :
[tex]\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \bigstar\:\:\bf Sum \:of \: zeroes\:\:: \end{gathered}[/tex]
[tex]\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} }\bigg\rgroup \\\\\end{gathered}[/tex]
[tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:}\\ \\\end{gathered}[/tex]
⠀⠀⠀⠀⠀⠀[tex]\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\end{gathered}[/tex]
[tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ Cofficient \:of \:x \:)\:}{Cofficient\:of\:x^2 \:} \\\\\end{gathered}[/tex]
[tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: -( \ 5 \:)\:}{ 1 \:} \\\\\end{gathered} [/tex]
[tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \dfrac{\: – 5 \:\:}{ 1 \:} \\\\\end{gathered}[/tex]
[tex]\begin{gathered}\qquad \dashrightarrow \:\sf p + q \: \: = \:\: \: – 5 \:\: \\\\\end{gathered}[/tex]
[tex]\begin{gathered}\qquad \dashrightarrow \underline{\pmb{\purple{\:p + q \: \: = \:\: \: – 5 \:\:\: }} }\:\bigstar \\\\\end{gathered}[/tex]
⠀⠀⠀⠀⠀[tex]\begin{gathered}\therefore {\underline{ \sf \:Hence,\:The\:value \:of\: p + q \:is\:\bf -5 }.}\\\\\end{gathered}[/tex]