Question

11. Find the zeroes of the quadratic polynomial 2x2 -x - 3 and verify the relationship
between the zeroes and the coefficients.​

Answer 1

Given :

• Quadratic polynomial 2x² - x - 3

To Find :

• Zeroes of the quadratic polynomial &
• Verify the relationship between zeroes and coefficients.

According to the question,

⇒ 2x² - x - 3

⇒ 2x² - 3x + 2x - 3

⇒ x(2x - 3) + 1(2x - 3)

⇒ (x + 1) (2x - 3)

Value of x,

⇒ x + 1 = 0

⇒ x = - 1

⇒ 2x - 3 = 0

⇒ x = 3/2

★ Verify the relationship between zeroes and the coefficients :-

★ Sum of zeroes :

⇒ α + β = -b/a

⇒ -1 + 3/2 = -(-1)/2

⇒ -2 + 3/2 = 1/2

⇒ 1/2 = 1/2

★ Product of zeroes :

⇒ αβ = c/a

⇒ (-1) × 3/2 = -3/2

⇒ -3/2 = -3/2

Hence, verified!

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Answer 2

Answer:

Given :-

• 2x² - x - 3

To Find :-

• What are the zeros of the quadratic polynomial and verify the relationship between the zeroes and co-efficient.

Solution :-

Given Equation :

$\bigstar \: \: \sf\bold{\purple{2x^2 - x - 3}}$

$\mapsto \sf 2x^2 - x - 3 =\: 0$

$\implies \sf 2x^2 - (3 - 2)x - 3 =\: 0$

$\implies \sf 2x^2 - 3x + 2x - 3 =\: 0$

$\implies \sf x(2x - 3) + 1(2x - 3) =\: 0$

$\implies \bf (x + 1) =\: 0$

$\longrightarrow \sf x + 1 =\: 0$

$\longrightarrow \sf\bold{\red{x =\: - 1}}$

$\implies \bf{(2x - 3) =\: 0}$

$\longrightarrow \sf 2x =\: 3$

$\longrightarrow \sf\bold{\red{x =\: \dfrac{3}{2}}}$

${\small{\bold{\underline{\therefore\: The\: zeroes\: are\: - 1\: and\: \dfrac{3}{2}\: respectively\: .}}}}$

Now, we have to verify the relationship between the zeroes and co-efficient ;

Given Equation :

$\bigstar\: \bf{2x^2 - x - 3}$

where,

• a = 2
• b = - 1
• c = - 3

➲ Sum of Roots :

As we know that :

$\mapsto \sf\boxed{\bold{\pink{Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

We have :

• a = 2
• b = - 1
• α = - 1
• β = 3/2

According to the question by using the formula we get,

$\implies \sf (- 1) + \dfrac{3}{2} =\: \dfrac{- (- 1)}{2}$

$\implies \sf - 1 + \dfrac{3}{2} =\: \dfrac{1}{2}$

$\implies \sf \dfrac{- 2 + 3}{2} =\: \dfrac{1}{2}$

$\implies \sf\bold{\green{\dfrac{1}{2} =\: \dfrac{1}{2}}}$

Hence, Verified.

➲ Product of Roots :

As we know that :

$\mapsto \sf\boxed{\bold{\pink{Product\: of\: roots\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

We have :

• a = 2
• c = - 3
• α = - 1
• β = 3/2

According to the question by using the formula we get,

$\implies \sf (- 1) \times \dfrac{3}{2} =\: \dfrac{- 3}{2}$

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

$\implies \sf \bold{\green{\dfrac{- 3}{2} =\: \dfrac{- 3}{2}}}$

Hence, Verified.