Question

find the zeros of the polynomial 6x2-3+7x and verify the relationship​

Answer 1

Step-by-step explanation:

Let 6x2 - 7x-3 be p(x) =0

so,6x2 - 7x-3 =0

6x2 - (9-2) x -3

6x2 - 9x + 2x -3

(6x2 - 9x)(2x-3)

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

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

⇒ 3x +1 = 0

⇒ 2x -3 = 0

⇒ x =3/2

⇒ x = - 1/3 .....

✌hope it helped u.... ✌

Answer 2

Question :-

Find the zeroes of the polynomial 6x² - 3 + 7x and verify the relationship between the zeroes and the coefficients .

Answer :-

Given :-

Quadratic equation : 6x² - 3 + 7x

Required to find :-

• Zeroes of the polynomial ?

• Relation between the zeroes and the coefficients ?

Solution :-

Quadratic equation : 6x² - 3 + 7x

The standard form of any quadratic equation is ;

• ax² + bx + c

Now,

Let's rearrange these terms of the quadratic equation in the such a way that it is in the standard form .

=> 6x² - 3 + 7x

=> 6x² + 7x - 3

Now,

Let's factorise this quadratic equation ;

=> 6x² + 7x - 3

=> 6x² + 9x - 2x - 3

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

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

This implies ;

2x + 3 = 0

2x = - 3

x = -3/2

Similarly,

3x - 1 = 0

3x = 1

x = 1/3

Hence,

The zeroes of the polynomial are -3/2 & 1/3

However,

Let's take this zeroes as ;

α = -3/2 , β = 1/3

Now,

Let's verify the relationship between the zeroes and the coefficients .

1st case

We know that ;

$\boxed{\tt{\orange{ \alpha + \beta = \dfrac{ - ( Coefficient \; of \; x ) }{ Coefficient \; of \; x^2 } }}}$

So,

α + β =

=> -3/2 + ( 1/3 )

=> - 3/2 + 1/3

=> - 9 + 2/6

=> - 7/6

But,

$\tt{ \dfrac{ - ( Coefficient \; of \; x )}{ Coefficient \; of \; x^2 } }$

=> - ( 7 )/6

=> -7/6

Hence,

$\boxed{\tt{ \alpha + \beta = \dfrac{ - ( Coefficient \; of \; x ) }{ Coefficient \; of \; x^2 } }}$

Similarly,

2nd case

We know that ;

$\green{\boxed{\tt{\red{ \alpha \beta = \dfrac{ Constant \; term }{ Coefficient \; of \; x^2 } }}}}$

α β =

=> -3/2 x 1/3

=> -1/2

But,

$\tt{ \dfrac{ Constant \; term }{ Coefficient \; of \; x^2 } }$

=> - 3/6

=> -1/2

Hence,

$\boxed{\tt{ \alpha \beta = \dfrac{ Constant \; term }{ Coefficient \; of \; x^2 } }}$

Therefore,

The relationship between the zeroes and the coefficients had been verified .