if one root of the polynomial p(y) = 5y² + 13y +m is reciprocal of other , then find the value of ‘m’ ? answer with solution… About the author Arya
EXPLANATION. Quadratic equation. ⇒ p(x) = 5y² + 13y + m. As we know that, Let one roots = α. Other roots is reciprocal = 1/α. Products of the zeroes of the quadratic equation. ⇒ αβ = c/a. ⇒ α x 1/α = m/5. ⇒ 1 = m/5. ⇒ m = 5. MORE INFORMATION. Nature of the roots of the quadratic expression. (1) = Real and unequal, if b² – 4ac > 0. (2) = Rational and different, if b² – 4ac is a perfect square. (3) = Real and equal, if b² – 4ac = 0. (4) = If D < 0 Roots are imaginary and unequal Or complex conjugate. Reply
Given that , The one root of the polynomial p(y) = 5y² + 13y +m is reciprocal of other . ❍ Let’s Consider the two roots of Polynomial be [tex]\bf \alpha \:[/tex] and [tex]\bf \beta \:[/tex] . The one root of the polynomial is reciprocal of other . [tex]\qquad\therefore \:\:\sf \beta \:=\: Reciprocal \: of \: \alpha \:\\\\[/tex] [tex]\qquad :\implies \sf \beta \:=\: Reciprocal \: of \: \alpha \: \:\\\\[/tex] [tex]\qquad :\implies \sf \beta \:=\: \dfrac{1}{\alpha } \: \:\\\\[/tex] [tex]\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\: \beta \:(\:or \: \:other \: root \:)\:=\: \dfrac{1}{\alpha } }}} }\:\:\bigstar \\[/tex] ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : p(y) = 5y² + 13 y + m ⠀⠀⠀⠀As , We know that , [tex]\qquad \underline { \boxed {\pmb{\red{\:\maltese \:\: Product \:of \: zeroes \:}\purple {\:\:(\:\alpha\:\beta \:)}\red{\::} }}}\\\\[/tex] [tex]\qquad \dashrightarrow \sf \:\:\alpha \:\: \beta \:\:=\:\:\dfrac{ \:Constant \:Term \:}{Cofficient \:of\:y^2 \:} \\\\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \dashrightarrow \sf \:\:\alpha \:\: \beta \:\:=\:\:\dfrac{ \:Constant \:Term \:}{Cofficient \:of\:y^2 \:} \\\\\qquad \dashrightarrow \sf \:\:\alpha \:\: \times \:\:\dfrac{1}{\alpha \:}\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:\cancel{\alpha} \:\: \times \:\:\dfrac{1}{\cancel {\alpha} \:}\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:1\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:1\:\times\: 5 \:=\:\: \:m \: \\\\\qquad \dashrightarrow \sf \:\:5 \:=\:\: \:m \: \\\\\qquad \dashrightarrow \sf \:\:m\:=\:\: \:5 \: \\\\\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\: \:\:m\:=\:\: \:5 \: }}} }\:\:\bigstar \\[/tex] [tex]\qquad \therefore \:\underline {\sf \:Hence,\:The \:value \:of \: m \: is \:\bf \:5 \:}\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ More To Know : [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
EXPLANATION.
Quadratic equation.
⇒ p(x) = 5y² + 13y + m.
As we know that,
Let one roots = α.
Other roots is reciprocal = 1/α.
Products of the zeroes of the quadratic equation.
⇒ αβ = c/a.
⇒ α x 1/α = m/5.
⇒ 1 = m/5.
⇒ m = 5.
MORE INFORMATION.
Nature of the roots of the quadratic expression.
(1) = Real and unequal, if b² – 4ac > 0.
(2) = Rational and different, if b² – 4ac is a perfect square.
(3) = Real and equal, if b² – 4ac = 0.
(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.
Given that , The one root of the polynomial p(y) = 5y² + 13y +m is reciprocal of other .
❍ Let’s Consider the two roots of Polynomial be [tex]\bf \alpha \:[/tex] and [tex]\bf \beta \:[/tex] .
[tex]\qquad\therefore \:\:\sf \beta \:=\: Reciprocal \: of \: \alpha \:\\\\[/tex]
[tex]\qquad :\implies \sf \beta \:=\: Reciprocal \: of \: \alpha \: \:\\\\[/tex]
[tex]\qquad :\implies \sf \beta \:=\: \dfrac{1}{\alpha } \: \:\\\\[/tex]
[tex]\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\: \beta \:(\:or \: \:other \: root \:)\:=\: \dfrac{1}{\alpha } }}} }\:\:\bigstar \\[/tex]
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : p(y) = 5y² + 13 y + m
⠀⠀⠀⠀As , We know that ,
[tex]\qquad \underline { \boxed {\pmb{\red{\:\maltese \:\: Product \:of \: zeroes \:}\purple {\:\:(\:\alpha\:\beta \:)}\red{\::} }}}\\\\[/tex]
[tex]\qquad \dashrightarrow \sf \:\:\alpha \:\: \beta \:\:=\:\:\dfrac{ \:Constant \:Term \:}{Cofficient \:of\:y^2 \:} \\\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \dashrightarrow \sf \:\:\alpha \:\: \beta \:\:=\:\:\dfrac{ \:Constant \:Term \:}{Cofficient \:of\:y^2 \:} \\\\\qquad \dashrightarrow \sf \:\:\alpha \:\: \times \:\:\dfrac{1}{\alpha \:}\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:\cancel{\alpha} \:\: \times \:\:\dfrac{1}{\cancel {\alpha} \:}\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:1\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:1\:\times\: 5 \:=\:\: \:m \: \\\\\qquad \dashrightarrow \sf \:\:5 \:=\:\: \:m \: \\\\\qquad \dashrightarrow \sf \:\:m\:=\:\: \:5 \: \\\\\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\: \:\:m\:=\:\: \:5 \: }}} }\:\:\bigstar \\[/tex]
[tex]\qquad \therefore \:\underline {\sf \:Hence,\:The \:value \:of \: m \: is \:\bf \:5 \:}\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ More To Know :
[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]