Question

Find the zeroes of 6x2

-3-7x and verify the relationship between the zeroes and the coefficients.​

Answer 1

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆⠀G I V E N ⠀ P O L Y N O M I A L : 6x² - 3 - 7x

⠀⠀⠀》⠀Finding out zeroes of Polynomial :

$\qquad \dashrightarrow \sf 6x^2 - 3 - 7x \\\\\qquad \dashrightarrow \sf 6x^2 - 7x - 3 \\\\\qquad \dashrightarrow \sf 6x^2 + 2x - 9x - 3 \\\\\qquad \dashrightarrow \sf 2x ( 3x + 1 )- 3 ( 3x + 1 ) \\\\\qquad \dashrightarrow \sf ( 2x - 3 ) ( 3x + 1 ) \\\\\qquad \dashrightarrow \sf x \:\:=\:\:\dfrac{3}{2}\:\:\:or\:\:x\:\:=\:\:-\dfrac{1}{3}\:\: \\\\\qquad \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:x \:\:=\:\:\:\:\dfrac{3}{2}\:\:\;or\:\:\:-\dfrac{1}{3}\:\:}}}}}$

$\qquad \therefore \:\:\underline {\sf\:Hence \:,\:The \:zeroes \:of \:Polynomial \:\:are \:\:\bf 3/2 \:\sf and \: \bf \:-1/3\:\sf.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀¤⠀Verifying relationship between zeroes and Cofficients :

$\qquad \underline {\boxed {\pmb{\bf { \:\maltese \:\:Sum\:\:of\:zeroes \:\purple { \:(\alpha \:+\:\beta \:)\:\:}\::}}}}$

$\dashrightarrow \sf \bigg( \:\:\alpha \:+\beta \:\bigg) \:\:=\:\:\dfrac{-(Cofficient \:of\:x\:)}{Cofficient \:of\:x^2\:}\:\\\\ \dashrightarrow \sf \bigg( \:\dfrac{3}{2}\:\:\bigg) \:+\bigg(\:-\dfrac{1}{3}` \:\bigg) \:\:=\:\:\dfrac{-(-7\:)}{\:6\:}\:\\\\\dashrightarrow \sf \bigg( \:\dfrac{3}{2}\:\: -\:\dfrac{1}{3}` \:\bigg) \:\:=\:\:\dfrac{-(-7\:)}{\:6\:}\:\\\\\dashrightarrow \sf \bigg( \:\dfrac{9 - 2 }{6}\:\: ` \:\bigg) \:\:=\:\:\dfrac{7\:}{\:6\:}\:\\\\\:\dashrightarrow \sf \bigg( \:\:\dfrac{7}{6}` \:\bigg) \:\:=\:\:\dfrac{7\:}{\:6\:}\:\\\\ \:\dashrightarrow \sf \dfrac{7}{6} \:\:=\:\:\dfrac{7}{\:6\:}\:\\\\ \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:\dfrac{7}{6}\:=\:\:\:\dfrac{7}{6}\:\:}}}}}$

⠀⠀⠀⠀⠀AND ,

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

$\dashrightarrow \sf \bigg( \:\:\alpha \:\beta \:\bigg) \:\:=\:\:\dfrac{\:Constant\:Term\:}{Cofficient \:of\:x^2\:}\:\\\\\dashrightarrow \sf \bigg( \:\dfrac{3}{2}\:\:\bigg) \:\bigg(\:-\dfrac{1}{3}` \:\bigg) \:\:=\:\:\dfrac{-3}{\:6\:}\:\\\\ \dashrightarrow \sf \:\dfrac{3}{2}\:\:\times \:\bigg(\:-\dfrac{1}{3} \:\bigg) \:\:=\:\:\dfrac{-3}{\:6\:}\:\\\\ \dashrightarrow \sf \: \:\bigg(\:\dfrac{-3}{6} \:\bigg) \:\:=\:\:\dfrac{-3}{\:6\:}\:\\\\ \dashrightarrow \sf \: \:\:\cancel{\dfrac{-3}{6}} \: \:\:=\:\:\cancel{\dfrac{-3}{\:6\:}}\:\\\\ \dashrightarrow \sf \: \:\:\dfrac{-1}{2} \: \:\:=\:\:\dfrac{-1}{\:2\:}\:\\\\ \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:\dfrac{-1}{2}\:=\:\:\:\dfrac{-1}{2}\:\:}}}}}$

⠀⠀⠀⠀⠀$\therefore {\underline {\pmb{\bf{ Hence, \:Verified \:}}}}$

Answer 2

Answer:

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆⠀G I V E N ⠀ P O L Y N O M I A L : 6x² - 3 - 7x

⠀⠀⠀》⠀Finding out zeroes of Polynomial :

$\begin{gathered}\qquad \dashrightarrow \sf 6x^2 - 3 - 7x \\\\\qquad \dashrightarrow \sf 6x^2 - 7x - 3 \\\\\qquad \dashrightarrow \sf 6x^2 + 2x - 9x - 3 \\\\\qquad \dashrightarrow \sf 2x ( 3x + 1 )- 3 ( 3x + 1 ) \\\\\qquad \dashrightarrow \sf ( 2x - 3 ) ( 3x + 1 ) \\\\\qquad \dashrightarrow \sf x \:\:=\:\:\dfrac{3}{2}\:\:\:or\:\:x\:\:=\:\:-\dfrac{1}{3}\:\: \\\\\qquad \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:x \:\:=\:\:\:\:\dfrac{3}{2}\:\:\;or\:\:\:-\dfrac{1}{3}\:\:}}}}}\\\\\end{gathered}$

⇢6x

2

−3−7x

⇢6x

2

−7x−3

⇢6x

2

+2x−9x−3

⇢2x(3x+1)−3(3x+1)

⇢(2x−3)(3x+1)

⇢x=

2

3

orx=−

3

1

⇢

x=

2

3

or−

3

1

x=

2

3

or−

3

1

$\begin{gathered}\qquad \therefore \:\:\underline {\sf\:Hence \:,\:The \:zeroes \:of \:Polynomial \:\:are \:\:\bf 3/2 \:\sf and \: \bf \:-1/3\:\sf.}\\\\\end{gathered}$

∴

Hence,ThezeroesofPolynomialare3/2and−1/3.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀¤⠀Verifying relationship between zeroes and Cofficients :

$\begin{gathered}\qquad \underline {\boxed {\pmb{\bf { \:\maltese \:\:Sum\:\:of\:zeroes \:\purple { \:(\alpha \:+\:\beta \:)\:\:}\::}}}}\\\\\end{gathered} </p><p>✠Sumofzeroes(α+β):</p><p>✠Sumofzeroes(α+β):$

$\begin{gathered} \dashrightarrow \sf \bigg( \:\:\alpha \:+\beta \:\bigg) \:\:=\:\:\dfrac{-(Cofficient \:of\:x\:)}{Cofficient \:of\:x^2\:}\:\\\\ \dashrightarrow \sf \bigg( \:\dfrac{3}{2}\:\:\bigg) \:+\bigg(\:-\dfrac{1}{3}` \:\bigg) \:\:=\:\:\dfrac{-(-7\:)}{\:6\:}\:\\\\\dashrightarrow \sf \bigg( \:\dfrac{3}{2}\:\: -\:\dfrac{1}{3}` \:\bigg) \:\:=\:\:\dfrac{-(-7\:)}{\:6\:}\:\\\\\dashrightarrow \sf \bigg( \:\dfrac{9 - 2 }{6}\:\: ` \:\bigg) \:\:=\:\:\dfrac{7\:}{\:6\:}\:\\\\\:\dashrightarrow \sf \bigg( \:\:\dfrac{7}{6}` \:\bigg) \:\:=\:\:\dfrac{7\:}{\:6\:}\:\\\\ \:\dashrightarrow \sf \dfrac{7}{6} \:\:=\:\:\dfrac{7}{\:6\:}\:\\\\ \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:\dfrac{7}{6}\:=\:\:\:\dfrac{7}{6}\:\:}}}}}\\\\\end{gathered}$

⇢(α+β)=

Cofficientofx

2

−(Cofficientofx)

⇢(

2

3

)+(−

3

1

‘)=

6

−(−7)

⇢(

2

3

−

3

1

‘)=

6

−(−7)

⇢(

6

9−2

‘)=

6

7

⇢(

6

7

‘)=

6

7

⇢

6

7

=

6

7

⇢

6

7

=

6

7

6

7

=

6

7

⠀⠀⠀⠀⠀AND ,

$\begin{gathered}\qquad \underline {\boxed {\pmb{\bf { \:\maltese \:\:Product \:\:of\:zeroes \:\purple { \:(\alpha \:\:\beta \:)\:\:}\::}}}}\\\\\end{gathered} </p><p>✠Productofzeroes(αβ):</p><p>✠Productofzeroes(αβ):$

$</p><p>\begin{gathered} \dashrightarrow \sf \bigg( \:\:\alpha \:\beta \:\bigg) \:\:=\:\:\dfrac{\:Constant\:Term\:}{Cofficient \:of\:x^2\:}\:\\\\\dashrightarrow \sf \bigg( \:\dfrac{3}{2}\:\:\bigg) \:\bigg(\:-\dfrac{1}{3}` \:\bigg) \:\:=\:\:\dfrac{-3}{\:6\:}\:\\\\ \dashrightarrow \sf \:\dfrac{3}{2}\:\:\times \:\bigg(\:-\dfrac{1}{3} \:\bigg) \:\:=\:\:\dfrac{-3}{\:6\:}\:\\\\ \dashrightarrow \sf \: \:\bigg(\:\dfrac{-3}{6} \:\bigg) \:\:=\:\:\dfrac{-3}{\:6\:}\:\\\\ \dashrightarrow \sf \: \:\:\cancel{\dfrac{-3}{6}} \: \:\:=\:\:\cancel{\dfrac{-3}{\:6\:}}\:\\\\ \dashrightarrow \sf \: \:\:\dfrac{-1}{2} \: \:\:=\:\:\dfrac{-1}{\:2\:}\:\\\\ \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:\dfrac{-1}{2}\:=\:\:\:\dfrac{-1}{2}\:\:}}}}}\\\\\end{gathered}$

⇢(αβ)=

Cofficientofx

2

ConstantTerm

⇢(

2

3

)(−

3

1

‘)=

6

−3

⇢

2

3

×(−

3

1

)=

6

−3

⇢(

6

−3

)=

6

−3

⇢

6

−3

=

6

−3

⇢

2

−1

=

2

−1

⇢

2

−1

=

2

−1

2

−1

=

2

−1

⠀⠀⠀⠀

$⠀\begin{gathered}\therefore {\underline {\pmb{\bf{ Hence, \:Verified \:}}}}\\\\\end{gathered}$

∴

Hence,Verified

Hence,Verified