Question

EXERCISE - 3.3
Find the zeroes of the following quadratic polynomials and verify the re
the zeroes and the coefficients
(iii) 6x? - 3-7x​

Answer 1

Question:-

Find the zeroes of the following polynomials and verify the relation between the zeroes and the coefficients.

Given polynomial:-

• (iii) 6x² - 7x - 3

Solution:-

Let us find the zeroes of the given polynomial by splitting the middle term.

= 6x² - 7x - 3

Splitting middle term,

= 6x² - 9x + 2x - 3

Taking common terms out,

⇒ 3x(2x - 3) + 1(2x - 3)

Taking (2x - 3) common,

⇒ (2x - 3)(3x + 1)

Either,

2x - 3 = 0

⇒ 2x = 3

⇒ x = 3/2

Or

3x + 1 = 0

⇒ 3x = -1

⇒ x = -1/3

∴ The zeroes of the given quadratic polynomial are 3/2 and -1/3

Now,

Let us verify the relation between the zeroes and the coefficients.

We know,

$\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

Hence,

$\sf{\dfrac{3}{2} + \dfrac{-1}{3} = \dfrac{-(-7)}{6}}$

$\sf{:\implies \dfrac{9 - 2}{6} = \dfrac{7}{6}}$

$\sf{:\implies \dfrac{7}{6} = \dfrac{7}{6}}$

Also,

$\sf{Product\:of\:zeroes = \dfrac{Constant\:Term}{Coefficient\:of\:x^2}}$

Hence,

$\sf{\dfrac{3}{2}\times \dfrac{-1}{3} = \dfrac{-3}{6}}$

$\sf{:\implies \dfrac{-1}{2} = \dfrac{-1}{2}}$

Hence, Verified!!!

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Answer 2

Answer:

Question:-

Find the zeroes of the following polynomials and verify the relation between the zeroes and the coefficients.

Given polynomial:-

(iii) 6x² - 7x - 3

Solution:-

Let us find the zeroes of the given polynomial by splitting the middle term.

= 6x² - 7x - 3

Splitting middle term,

= 6x² - 9x + 2x - 3

Taking common terms out,

⇒ 3x(2x - 3) + 1(2x - 3)

Taking (2x - 3) common,

⇒ (2x - 3)(3x + 1)

Either,

2x - 3 = 0

⇒ 2x = 3

⇒ x = 3/2

Or

3x + 1 = 0

⇒ 3x = -1

⇒ x = -1/3

∴ The zeroes of the given quadratic polynomial are 3/2 and -1/3

Now,

Let us verify the relation between the zeroes and the coefficients.

We know,

\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}Sumofzeroes=

Coefficientofx

2

−Coefficientofx

Hence,

\sf{\dfrac{3}{2} + \dfrac{-1}{3} = \dfrac{-(-7)}{6}}

2

3

+

3

−1

=

6

−(−7)

\sf{:\implies \dfrac{9 - 2}{6} = \dfrac{7}{6}}:⟹

6

9−2

=

6

7

\sf{:\implies \dfrac{7}{6} = \dfrac{7}{6}}:⟹

6

7

=

6

7

Also,

\sf{Product\:of\:zeroes = \dfrac{Constant\:Term}{Coefficient\:of\:x^2}}Productofzeroes=

Coefficientofx

2

ConstantTerm

Hence,

\sf{\dfrac{3}{2}\times \dfrac{-1}{3} = \dfrac{-3}{6}}

2

3

×

3

−1

=

6

−3

\sf{:\implies \dfrac{-1}{2} = \dfrac{-1}{2}}:⟹

2

−1

=

2

−1

Hence, Verified!!!

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