Question

Find the zeroos of following quadratic polynomials
and verify the relationship between the zeroes
and Co officients
x^2-6x+9

Answer 1

Step-by-step explanation:

Given:

Polynomial = x² - 6x + 9

To find:

Verifying the relationship between zeros of polynomial and it's co-officients.

Solution:

Find zeros of x² - 6x + 9

$\sf{\longrightarrow{{x}^{2} - 6x + 9}}$

$\sf{\longrightarrow{ {x}^{2} - 3x - 3x + 9}}$

$\sf{\longrightarrow{x(x - 3) - 3(x - 3)}}$

$\sf{\longrightarrow{(x - 3)(x - 3) = 0}}$

$\sf{\longrightarrow} \: x = 3 \: and \: x = 3$

The zeros are 3 and 3

_____________________

Verifying the relationship:

For polynomial x² - 6x + 9, if zeros are α and β

In the polynomial -

• a = 1
• b = -6
• c = 9

Sum of zeros :

$\sf{\longrightarrow} \: \alpha + \beta = \dfrac{ - b}{a}$

$\sf{\longrightarrow} \: \alpha + \beta = \dfrac{ - ( - 6)}{1}$

$\sf{\longrightarrow} \: \alpha + \beta = 6$

Sum of the zeros is 6.

_____________________

Product of zeros :

$\sf{\longrightarrow} \: \alpha \times \beta = \dfrac{c}{a}$

$\sf{\longrightarrow} \: \alpha \times \beta = \dfrac{9}{1}$

$\sf{\longrightarrow} \: \alpha \times \beta = 9$

Product of zeros = 9

Hence verified!

Answer 2

Answer:

Given :-

$\mapsto \sf {x}^{2} - 6x + 9$

To Find :-

• What is the zeros and verify the relationship between the zeros and it's co-efficient.

Solution :-

Given equation :

${\large{\bold{\purple{\underline{\rightarrow\: {x}^{2} - 6x + 9}}}}}$

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

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

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

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

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

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

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

And,

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

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

$\sf \therefore The\: two\: roots\: is\: \alpha =\: 3\: and\: \beta =\: 3.$

$\rule{300}{2}$

$\diamondsuit \: \boxed{\sf{VERIFY\: THE\: RELATIONSHIP\: BETWEEN\: ZEROS\: AND\: CO-EFFICIENT\: :-}}$

Let, the two roots of the polynomial be α and β

$\longmapsto$ Sum of roots :

$\clubsuit \: \sf\boxed{\bold{\pink{Sum\: of\: roots =\: \dfrac{- b}{a}}}}$

Given :

• α = 3
• β = 3

$\leadsto$ $\sf\bold{\blue{{x}^{2} - 6x + 9}}$

where,

• a = 1
• b = - 6
• c = 9

So,

$\implies \sf \alpha + \beta =\: \dfrac{- b}{a}$

$\implies \sf 3 + 3 =\: \dfrac{- (- 6)}{1}$

$\implies \sf 6 =\: \dfrac{6}{1}$

$\implies \sf\bold{\purple{6 =\: 6}}$

Again,

$\longmapsto$ Product of roots :

$\clubsuit \: \sf\boxed{\bold{\pink{Product\: of\: roots =\: \dfrac{c}{a}}}}$

Given :

• α = 3
• β = 3

$\leadsto \sf\bold{\blue{ {x}^{2} - 6x + 9}}$

where,

• a = 1
• b = - 6
• c = 9

So,

$\implies \sf \alpha\beta =\: \dfrac{c}{a}$

$\implies \sf 3 \times 3 =\: \dfrac{9}{1}$

$\implies \sf\bold{\purple{9 =\: 9}}$

$\longmapsto$ HENCE, VERIFIED .