Question

YNOMIALS 33

EXERCISE 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between

the zeroes and the coefficients.

(i) x2

– 2x – 8 (ii) 4s2

– 4s + 1 (iii) 6x2

– 3 – 7x

(iv) 4u2

+ 8u (v) t

2

– 15 (vi) 3x2

– x – 4

2. Find a quadratic polynomial each with the given numbers as the sum and product of its

zeroes ​

Answer 1

Answer:

first of all try yourself by help of examples dont be dependent on anyone in your exam no one will help you if then also you are anable to solve then ask me again and again till you do not understand how to solve pls ignore the above statement if you have tried and sorry if you tried to solve

Answer 2

Formula Used :-

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

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

Solution :-

${\small{\bold{\green{\underline{i)\: {x}^{2} - 2x - 8}}}}}$

$\mapsto \sf f(x) =\: {x}^{2} - 2x - 8$

$\implies \sf {x}^{2} - (4 - 2)x - 8 =\: 0$

$\implies \sf {x}^{2} - 4x + 2x - 8 =\: 0$

$\implies \sf x(x - 4) + 2(x - 4) =\: 0$

$\implies \sf (x + 2)(x - 4) =\: 0$

$\implies \sf x + 2 =\: 0$

$\implies \sf\bold{\red{x =\: - 2}}$

And,

$\implies \sf x - 4 =\: 0$

$\implies \sf\bold{\red{x =\: 4}}$

$\therefore$ The zeroes of the polynomial is - 2 and 4.

$\clubsuit$ VERIFICATION

Given equation :

$\mapsto$ x² - 2x - 8

where,

a = 1

b = - 2

c = - 8

$\diamondsuit$ Sum of the zeroes :

$\leadsto \sf - 2 + 4 =\: \dfrac{- (- 2)}{1}$

$\leadsto \sf - 2 + 4 =\: \dfrac{2}{1}$

$\leadsto\sf\bold{\purple{2 =\: 2}}$

Again,

$\diamondsuit$ Product of the zeroes :

Then,

$\leadsto \sf - 2 \times 4 =\: \dfrac{- 8}{1}$

$\leadsto\sf\bold{\purple{- 8 =\: - 8}}$

Hence, Verified.

$\rule{150}{2}$

${\small{\bold{\green{\underline{ii)\: 4{s}^{2} - 4s + 1}}}}}$

$\mapsto\sf f(x) =\: 4{s}^{2} - 4s + 1$

$\implies \sf 4{s}^{2} - (2 + 2)s + 1 =\: 0$

$\implies \sf 4{s}^{2} - 2s + 2s + 1 =\: 0$

$\implies \sf 2s(2s - 1) - 1(2s - 1) =\: 0$

$\implies \sf (2s - 1) (2s - 1) =\: 0$

$\implies \sf 2s - 1 =\: 0$

$\implies \sf 2s =\: 1$

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

And,

$\implies\sf 2s - 1 =\: 0$

$\implies \sf 2s =\: 1$

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

$\therefore$ The zeroes of the polynomial is ½ and ½.

$\clubsuit$ VERIFICATION

Given equation :

$\mapsto$ 4s² - 4s + 1

where,

a = 4

b = - 4

c = 1

$\diamondsuit$ Sum of the zeroes :

$\leadsto \sf \dfrac{1}{2} + \dfrac{1}{2} =\: \dfrac{- (- 4)}{4}$

$\leadsto\sf \dfrac{1}{2} + \dfrac{1}{2} =\: \dfrac{4}{4}$

$\leadsto\sf \dfrac{2 + 2}{4} =\: \dfrac{4}{4}$

$\leadsto \sf \dfrac{\cancel{4}}{\cancel{4}} =\: \dfrac{\cancel{4}}{\cancel{4}}$

$\leadsto\sf\bold{\purple{1 =\: 1}}$

Again,

$\diamondsuit$ Product of the zeroes :

$\leadsto \sf \dfrac{1}{2} \times \dfrac{1}{2} =\: \dfrac{1}{4}$

$\leadsto\sf\bold{\red{\dfrac{1}{4} =\: \dfrac{1}{4}}}$

Hence Verified.

$\rule{150}{2}$

${\small{\bold{\green{\underline{iii)\: 6{x}^{2} - 7x - 3}}}}}$

$\mapsto \sf f(x) =\: 6{x}^{2} - 7x - 3$

$\implies \sf 6{x}^{2} - (9 - 2)x - 3 =\: 0$

$\implies \sf 6{x}^{2} - 9x + 2x - 3 =\: 0$

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

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

$\implies\sf 3x + 1 =\: 0$

$\implies \sf 3x =\: - 1$

$\implies \sf\bold{\red{x =\: \dfrac{- 1}{3}}}$

And,

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

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

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

$\therefore$ The zeroes of the polynomial is - 1/3 and 3/2.

$\clubsuit$ VERIFICATION

Given equation :

$\mapato$ 6x² - 7x - 3

where,

a = 6

b = - 7

c = - 3

$\diamondsuit$ Sum of the zeroes :

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

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

$\leadsto \sf\bold{\purple{\dfrac{7}{6} =\: \dfrac{7}{6}}}$

Again,

$\diamondsuit$ Product of the zeroes :

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

$\leadsto \sf \dfrac{- 3}{6} =\: \dfrac{- 3}{6}$

$\leadsto\sf \dfrac{-\cancel{3}}{\cancel{6}} =\: \dfrac{-\cancel{3}}{\cancel{6}}$

$\leadsto\sf\bold{\purple{\dfrac{-1}{2} =\: \dfrac{- 1}{2}}}$

Hence, Verified.

$\rule{150}{2}$

iv) 3x² - x - 4

↦ f(x) = 3x² - x - 4

⇒ 3x² - (4 - 3)x - 4 = 0

⇒ 3x² - 4x + 3x - 4 = 0

⇒ x(3x - 4) + 1(3x - 4) = 0

⇒ (3x - 4)(x + 1) = 0

⇒ 3x - 4 = 0

⇒ 3x = 4

➠ x = 4/3

And

⇒ x + 1 = 0

➠ x = - 1

∴ The zeroes of the polynomial is 4/3 and - 1

✪ VERIFICATION

Given equation :

➲ 3x² - x - 4

where,

a = 3

b = - 1

c = - 4

★ Sum of the zeroes :

⇒ 4/3 + - 1 = - (- 1)/3

⇒ 4 - 3/3 = 1/3

➦ 1/3 = 1/3

Again,

★ Product of the zeroes :

⇒ 4/3 × (- 1) = - 4/3

➦ - 4/3 = - 4/3

Hence, Verified.

$\rule{150}{2}$

v) t² - 15

↦ f(x) = t² - 15

⇒ t² - 15 = 0

⇒ (t)² - (√15)² = 0

⇒ t² - (√15)² = 0

⇒ (t + √15) (t - √15) = 0

⇒ t + √15 = 0

➠ t = - √15

⇒ t - √15 = 0

➠ t = √15

∴ The zeroes of the polynomial is √15 and - √15.

✪ VERIFICATION

Given equation :

➲ t² - 15

where,

a = 1

b = 0

c = - 15

★ Sum of the zeroes :

⇒ √15 + (- √15) = - (0/1)

➦ 0 = 0

★ Product of the zeroes :

⇒ √15 × (- √15) = - 15/1

➦ - 15 = - 15

Hence, Verified.

[Note : Apologize for not using the latex for the problem no iv and v. Because we can't answer a question more than 5000 words. ]