Question

If α and β are the zeroes of the polynomial x2 + 3x – 4, find a quadratic polynomial whose zeroes are 1/α and 1/β.

Answer 1

Given that , α and β are the zeroes of the polynomial x² + 3x – 4 .

Need To Find : A quadratic polynomial whose zeroes are 1/α and 1/β ? ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \pmb{\mathbb{ POLYNOMIAL \:\::\:\: } \sf x^2 + 3x - 4 \:}$

As , We know that,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Sum \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha + \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \bigg( \: \alpha \:+\:\beta \:\bigg) \:\:=\:\:\dfrac{\:-\:(\:Cofficient\:of \:x\:)}{Cofficient \:of \: x^2 \:}\\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \bigg( \: \alpha \:+\:\beta \:\bigg) \:\:=\:\:\dfrac{\:-\:(\:Cofficient\:of \:x\:)}{Cofficient \:of \: x^2 \:}\\\\\dashrightarrow \sf \bigg( \: \alpha \:+\:\beta \:\bigg) \:\:=\:\:\dfrac{\:-\:(\:3\:)}{\:1 \:}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:\dfrac{\:-\:3\:}{\:1 \:}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:-3 \:\\\\\dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha \:+\:\beta \: \:\:=\:\:-3 \:\:}}}}}\:\:\bigstar$

⠀AND ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Product \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \bigg( \: \alpha \:\:\beta \:\bigg) \:\:=\:\:\dfrac{Constant \:Term }{Cofficient \:of \: x^2 \:}\\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \bigg( \: \alpha \:\:\beta \:\bigg) \:\:=\:\:\dfrac{\:Constant \:Term \:}{Cofficient \:of \: x^2 \:}\\\\\dashrightarrow \sf \bigg( \: \alpha \:\:\beta \:\bigg) \:\:=\:\:\dfrac{\:-4}{\:1 \:}\\\\\dashrightarrow \sf \: \alpha \:\:\beta \: \:\:=\:\:\dfrac{\:-\:4\:}{\:1 \:}\\\\\dashrightarrow \sf \: \alpha \:\:\beta \: \:\:=\:\:-4 \:\\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha \:\:\beta \: \:\:=\:\:-4 \:\:}}}}}\:\:\bigstar$⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀¤ Finding quadratic polynomial whose zeroes are 1/α and 1/β :

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Sum \:\: of \:\:zeroes \:\:\purple{ \:(\: 1 /\alpha + 1/\beta \;)\:} \:: \: }}}\\\\ \dashrightarrow \sf \dfrac{1}{\alpha} + \dfrac{1}{\beta} \\\\\dashrightarrow \sf \dfrac{( \alpha + \beta\:) }{\alpha \beta }$

Here ,

• α + β = -3
• α β = -4

$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\dashrightarrow \sf \dfrac{( \alpha + \beta\:) }{\alpha \beta } \\\\\dashrightarrow \sf \dfrac{( -3\:) }{-4 } \\\\\dashrightarrow \sf \dfrac{3 }{4 } \\\\\dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \dfrac{1}{\alpha} + \:\:\dfrac{1}{\beta} \: \:\:=\:\:\dfrac{3}{4} \:\:}}}}}\:\:\bigstar$⠀⠀

AND ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Product \:\: of \:\:zeroes \:\:\purple{ \:(\: 1 /\alpha 1/\beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \dfrac{1}{\alpha} \times \dfrac{1}{\beta} \\\\\dashrightarrow \sf \dfrac{1 }{\alpha \beta }$

As , We know that ,

• α β = -4

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \dfrac{( 1\:) }{\alpha \beta } \\\\\dashrightarrow \sf \dfrac{1 }{-4 } \\\\ \dashrightarrow \sf \dfrac{-1 }{4 } \\\\\dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \dfrac{1}{\alpha} \:\:\dfrac{1}{\beta} \: \:\:=\:\:\dfrac{-1}{4} \:\:}}}}}\:\:\bigstar$

As , We know that ,

$\qquad \star \:\:\underline {\boxed {\pink{\pmb{\frak{ Quadratic \:Equation \:=\:x^2 \: - \: ( Sum \:of \:zeroes \:)x \: +\:Product \:of \:zeroes \: \:=0\:\:}}}}}$

Therefore ,

$\qquad \underline {\bigstar \:\:\purple { \bf Quadratic \:Polynomial \:whose \: zeroes \:are \:1/\alpha \:\& \:1/\beta \:\::\:}}\\\\\qquad \dashrightarrow \sf \:x^2 \: - \: ( Sum \:of \:zeroes \:)x \: +\:Product \:of \:zeroes \: = 0 \\\\\qquad \dashrightarrow \sf \:x^2 \: - \: \bigg( \: \dfrac{1}{\alpha} + \:\:\dfrac{1}{\beta} \:\bigg)x \: +\:\bigg( \: \dfrac{1}{\alpha} \:\:\dfrac{1}{\beta} \:\bigg) \: = 0$

Here ,

• 1/α + 1/β = 3/4
• 1/α × 1/β = -1/4

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\qquad \dashrightarrow \sf \:x^2 \: - \: \bigg( \: \dfrac{1}{\alpha} + \:\:\dfrac{1}{\beta} \:\bigg)x \: +\:\bigg( \: \dfrac{1}{\alpha} \:\:\dfrac{1}{\beta} \:\bigg) \: = 0\\\\\qquad \dashrightarrow \sf \:x^2 \: - \: \bigg( \: \dfrac{3}{4} \:\bigg)x \: +\:\bigg( \: \dfrac{-1}{4} \:\bigg) \: = 0\\\\\qquad \dashrightarrow \sf \:x^2 \: - \: \: \dfrac{3}{4}x \: - \: \dfrac{1}{4} \: \:= 0 \\\\\qquad \dashrightarrow \sf \:x^2 \: - \: \: \dfrac{3}{4}x \: - \: \dfrac{1}{4} \:= \:0 \\\\\qquad \dashrightarrow \sf \dfrac{\:4x^2 \: - \: \: 3x \: -1}{4} \:= \:0 \\\\\qquad \dashrightarrow \sf \:4x^2 \: - \: \: 3x \: -1 \:= \:0$

$\qquad \therefore \:\underline {\sf Hence, \:The \:Quadratic \:Equation \:whose \:zeroes \:are \:1/\alpha \:\& \:1/\beta \:are \:\pmb{\bf 4x^2 - 3x - 1 \:}\:.}$