Question

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) t2-15
(vi) 3x2 -x-4​

Answer 1

Step-by-step explanation:

(1) x² - 2x -8

x² - 4x + 2x - 8

x(x-4).+2(x-4)

(x+2)(x-4)=0

x = -2 and 4

verification

sum of zeroes(α+β) = -b/a

-2 + 4 = -(-2)/1

2 = 2

product of zeroes (αβ) = c/a

-2 × 4 = -8/1

-8 = -8

(2) 4s² - 4s + 1

4s² -2s -2s +1

2s(2s - 1).-1(2s - 1)

(2s - 1)(2s - 1)

s = 1/2 and 1/2

verification

α+β = -b/a

1/2 + 1/2 = -(-4)/4

2/2 = 4/4

1 = 1

αβ = c/a

1/2 × 1/2 = 1/4

1/4 = 1/4

(3) 6x² -7x -3

6x² - 9x + 2x - 3

3x(2x-3).+1(2x-3)

(3x + 1)(2x-3) = 0

x = -1/3 and 3/2

verification

α + β = -b/a

-1/3 + 3/2 = -(-7)/6

(-2+9)/6 = 7/6

7/6 = 7/6

αβ = c/a

-1/3 × 3/2 = -3/6

-1/2 = -1/2

(4) 4u² + 8u

4u(u+2) = 0

u = 0 and -2

verification

α + β = -b/a

o + (-2) = -8/4

-2 = -2

αβ = c/a

0 × (-2) = 0/4

0 = 0

(5) t² - 15 = 0

t² = 15

t = ±√15

t = +√15 , -√15

verification

α + β = -b/a

-√15 + √15 = 0/1

0 = 0

αβ = c/a

(-√15) × (√15) = -15/1

-15 = -15

(6) 3x² - x - 4

3x² -4x + 3x - 4

x(3x-4).+1(3x-4)

(x+1)(3x-4) = 0

x = -1 , 4/3

verification

α + β = -b/a

-1 + 4/3 = -(-1)/3

(-3+4)/3 = 1/3

1/3 = 1/3

αβ = c/a

-1 × 4/3 = -4/3

-4/3 = -4/3

Answer 2

$\rm{\color{blue}{❤||Mehak\:||❤}}$

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:-