Question

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POLYNOMIALS
pag no:- 33
EXERCISE 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(i) 4S2 – 4s + 1
(ii) 6x2-3-7x
(iii) 4u2+8u
(iv) t2-15
(v) 3x2-x-4
​

Answer 1

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. ]