Question

find the zeros of the quadratic polynomial 6x2+29 x+35 and verify the relationship between the zeroes and the coefficient of the polynomial​

Answer 1

$\large {\red {\underline {\overline {\mid {\sf {Aim:-{\mid }}}}}}}$

To find the zeroes of the quadratic polynomial $6x² - 29x +35$ and verify the relationship between the zeroes and the coefficient of the polynomial.

$\large {\red {\underline {\overline {\mid {\sf {Prerequisite \: Concept:-{\mid }}}}}}}$

• $\small{\orange {\boxed {\bf {Factorization \: method}}}}$
• $\small{\orange {\boxed {\bf {Sum\: of\: zeroes = \frac{-b}{a}}}}}$
• $\small{\orange {\boxed {\bf {Product \: of \: zeroes = \frac{c}{a}}}}}$

$\large {\red {\underline {\overline {\mid {\sf {Calculation:-{\mid }}}}}}}$

$Factorizing \: the\: equation, \\ ⇒6x² - 29x + 35 = 0\\ ⇒6x² - 14x - 15x + 35 = 0 \\⇒2x(3x - 7) -5(3x - 7) =0\\ ⇒(2x - 5)(3x - 7) = 0 \\ ⇒x = \frac{5}{2}, \frac{7}{3}$

$Sum\: of \:zeroes = \frac{5}{2} + \frac{7}{3} = \frac{29}{6} \\ Sum \: of \: zeroes,\: by \: coefficients = \frac{-b}{a} = \frac{29}{6} \\ \\ Product \: of \: zeroes = \frac{5}{2} × \frac{7}{3}= \frac{35}{6} \\ Product\: of \: zeroes,\: by \: coefficients = \frac{c}{a} = \frac{35}{6}\\ Hence, \: verified$

Answer 2

Answer:

\large {\red {\underline {\overline {\mid {\sf {Aim:-{\mid }}}}}}}

∣Aim:−∣

To find the zeroes of the quadratic polynomial 6x² - 29x +356x²−29x+35 and verify the relationship between the zeroes and the coefficient of the polynomial.

\large {\red {\underline {\overline {\mid {\sf {Prerequisite \: Concept:-{\mid }}}}}}}

∣PrerequisiteConcept:−∣

\small{\orange {\boxed {\bf {Factorization \: method}}}}

Factorizationmethod

\small{\orange {\boxed {\bf {Sum\: of\: zeroes = \frac{-b}{a}}}}}

Sumofzeroes=

a

−b

\small{\orange {\boxed {\bf {Product \: of \: zeroes = \frac{c}{a}}}}}

Productofzeroes=

a

c

\large {\red {\underline {\overline {\mid {\sf {Calculation:-{\mid }}}}}}}

∣Calculation:−∣

\begin{gathered} Factorizing \: the\: equation, \\ ⇒6x² - 29x + 35 = 0\\ ⇒6x² - 14x - 15x + 35 = 0 \\⇒2x(3x - 7) -5(3x - 7) =0\\ ⇒(2x - 5)(3x - 7) = 0 \\ ⇒x = \frac{5}{2}, \frac{7}{3} \end{gathered}

Factorizingtheequation,

⇒6x²−29x+35=0

⇒6x²−14x−15x+35=0

⇒2x(3x−7)−5(3x−7)=0

⇒(2x−5)(3x−7)=0

⇒x=

2

5

,

3

7

\begin{gathered} Sum\: of \:zeroes = \frac{5}{2} + \frac{7}{3} = \frac{29}{6} \\ Sum \: of \: zeroes,\: by \: coefficients = \frac{-b}{a} = \frac{29}{6} \\ \\ Product \: of \: zeroes = \frac{5}{2} × \frac{7}{3}= \frac{35}{6} \\ Product\: of \: zeroes,\: by \: coefficients = \frac{c}{a} = \frac{35}{6}\\ Hence, \: verified \end{gathered}

Sumofzeroes=

2

5

+

3

7

=

6

29

Sumofzeroes,bycoefficients=

a

−b

=

6

29

Productofzeroes=

2

5

×

3

7

=

6

35

Productofzeroes,bycoefficients=

a

c

=

6

35

Hence,verified