Question

Find the zero of quadratic polynomial and verify the relation between zeros and it's coefficient 1. 4x2 - 9 2. X2 - 2x - 8

Answer 1

Given : -

Quadratic polynomials ;

• 1. 4x² - 9

• 2. x² - 2x - 8

Required to find : -

• Zeroes of the quadratic polynomial ?

• Relation between the zeroes and the coefficients ?

Solution : -

1.

4x² - 9

Let's factorise this polynomial .

=> 4x² - 9

=> ( 2x )² - ( 3 )²

This is in the form of an identity ;

a² - b² = ( a + b ) ( a - b )

This implies ;

=> ( 2x )² - ( 3 )²

=> ( 2x + 3 ) ( 2x - 3 )

2x + 3 = 0

2x = - 3

• x = - 3/2

2x - 3 = 0

2x = 3

• x = 3/2

Hence,

- 3/2 & 3/2 are the zeroes of the polynomial .

Now,

Let's verify the relationship ;

Firstly ,

Let's consider these 2 zeroes are ;

α = - 3/2 , β = 3/2

This implies ;

α + β = - coefficient of x/ coefficient of x²

=> - 3/2 + 3/2 = - 0/4

=> - 3+3/2 = 0/4

=> 0/2 = 0

=> 0 = 0

=> LHS = RHS

Similarly,

α β = constant term/ coefficient of x²

=> - 3/2 x 3/2 = -9/4

=> -9/4 = -9/4

=> LHS = RHS

Hence,

The relationship between the zeroes and coefficients is verified !

$\rule{400}{2}$

2.

x² - 2x - 8

Let's factorise this polynomial .

=> x² - 2x - 8

The middle term can be splited as ;

- 4x , + 2x because ,

The sum of these 2 terms is equal to the middle term and the product is also equal to the product of the coefficient of x² & constant term .

This implies :

=> x² - 4x + 2x - 8

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

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

x - 4 = 0

• x = 4

x + 2 = 0

• x = - 2.

Hence,

4 , - 2 are the zeroes of the Polynomial .

Now,

Let's verify the relationship ;

Firstly ,

Let's consider these 2 zeroes are ;

α = 4 , β = - 2

This implies ;

α + β = - coefficient of x/ coefficient of x²

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

=> 4 - 2 = 2/1

=> 2 = 2

=> LHS = RHS

Similarly,

α β = constant term/ coefficient of x²

=> 4 x - 2 = - 8/1

=> - 8 = -8

=> LHS = RHS

Hence,

The relationship between the zeroes and coefficients is verified !

Answer 2

$\mathcal{\huge{\fbox{\red{Question:-}}}}$

✴ Find the zero of quadratic polynomial and verify the relation between zeros and it's coefficient.

1.) 4x² - 9

2.) x² - 2x - 8

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

⚽ Given :-

• Quadratic polynomials ;

1. 4x² - 9

2. x² - 2x - 8

✨ To find :-

• Zeroes of the quadratic polynomial ?

• Relation between the zeroes and the coefficients ?

⭕ Solution : -

1.) 4x² - 9

=> 4x² - 9

=> ( 2x )² - ( 3 )²

Using, a² - b² where, ( a² - b² )( a + b ) ( a - b )

=> ( 2x )² - ( 3 )²

=> ( 2x + 3 ) ( 2x - 3 )

=> 2x + 3 = 0

=> 2x = - 3

=> x = - 3/2

=> 2x - 3 = 0

=> 2x = 3

=> x = 3/2

Hence, - 3/2 & 3/2 are the zeroes of the polynomial .

Verification :-

α = - 3/2 , β = 3/2

α + β = - coefficient of x/ coefficient of x²

=> - 3/2 + 3/2 = - 0/4

=> - 3+3/2 = 0/4

=> 0/2 = 0

=> 0 = 0

=> LHS = RHS

&

α β = constant term/ coefficient of x²

=> - 3/2 x 3/2 = -9/4

=> -9/4 = -9/4

=> LHS = RHS

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2.) x² - 2x - 8

=> x² - 2x - 8

Using, The middle term Factorisation,

=> x² - 4x + 2x - 8

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

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

=> x - 4 = 0

=> x = 4

x + 2 = 0

=> x = - 2.

Hence, 4 , - 2 are the zeroes of the Polynomial .

Verification :-

Here, α = 4 , β = - 2

α + β = - coefficient of x/ coefficient of x²

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

=> 4 - 2 = 2/1

=> 2 = 2

=> LHS = RHS

&

α β = constant term/ coefficient of x²

=> 4 x - 2 = - 8/1

=> - 8 = -8

=> LHS = RHS

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