Question

if one root of the polynomial p(y) = 5y² + 13y +m is reciprocal of other , then find the value of 'm' ?

answer with solution...
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Answer 1

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.

Answer 2

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 $\bf \alpha \:$ and $\bf \beta \:$ .

• The one root of the polynomial is reciprocal of other .

$\qquad\therefore \:\:\sf \beta \:=\: Reciprocal \: of \: \alpha \:$

$\qquad :\implies \sf \beta \:=\: Reciprocal \: of \: \alpha \: \:$

$\qquad :\implies \sf \beta \:=\: \dfrac{1}{\alpha } \: \:$

$\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\: \beta \:(\:or \: \:other \: root \:)\:=\: \dfrac{1}{\alpha } }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : p(y) = 5y² + 13 y + m

⠀⠀⠀⠀As , We know that ,

$\qquad \underline { \boxed {\pmb{\red{\:\maltese \:\: Product \:of \: zeroes \:}\purple {\:\:(\:\alpha\:\beta \:)}\red{\::} }}}$

$\qquad \dashrightarrow \sf \:\:\alpha \:\: \beta \:\:=\:\:\dfrac{ \:Constant \:Term \:}{Cofficient \:of\:y^2 \:}$

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

$\qquad \dashrightarrow \sf \:\:\alpha \:\: \beta \:\:=\:\:\dfrac{ \:Constant \:Term \:}{Cofficient \:of\:y^2 \:} \\\\\qquad \dashrightarrow \sf \:\:\alpha \:\: \times \:\:\dfrac{1}{\alpha \:}\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:\cancel{\alpha} \:\: \times \:\:\dfrac{1}{\cancel {\alpha} \:}\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:1\:\:=\:\:\dfrac{ \:m \:}{5\:} \\\\\qquad \dashrightarrow \sf \:\:1\:\times\: 5 \:=\:\: \:m \: \\\\\qquad \dashrightarrow \sf \:\:5 \:=\:\: \:m \: \\\\\qquad \dashrightarrow \sf \:\:m\:=\:\: \:5 \: \\\\\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\: \:\:m\:=\:\: \:5 \: }}} }\:\:\bigstar$

$\qquad \therefore \:\underline {\sf \:Hence,\:The \:value \:of \: m \: is \:\bf \:5 \:}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ More To Know :

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