Question

Need help on this question because I have pre-board of maths day after tomorrow, need help please.

Thanks.​

Answer 1

Question:

If the zeroes of the quadratic polynomial x² + ( a + 1 ) x + b are 2 and - 3, then

a) a = - 7, b = - 1

b) a = 5, b = - 1

c) a = 2, b = - 6

d) a = 0, b = - 6

Answer:

The values of a and b are ( a, b ) = ( 0, - 6 ).

Option d) a = 0, b = - 6

Step-by-step-explanation:

The given quadratic polynomial is x² + ( a + 1 ) x + b.

We have given that,

The zeroes of the quadratic polynomial are 2 & - 3.

By substituting x = 2 in the given quadratic polynomial, we get 0.

x² + ( a + 1 ) x + b

⇒ ( 2 )² + ( a + 1 ) * 2 + b = 0

⇒ 4 + 2a + 2 + b = 0

⇒ 2a + b + 6 = 0

⇒ 2a + b = - 6

⇒ b = - 6 - 2a

⇒ b = - 2a - 6 - - - ( 1 )

Now,

By substituting x = - 3 in the given quadratic polynomial, we get 0.

x² + ( a + 1 ) x + b

⇒ ( - 3 )² + ( a + 1 ) * ( - 3 ) + b = 0

⇒ 9 - 3a - 3 + b = 0

⇒ - 3a + b + 6 = 0

⇒ - 3a + b = - 6

⇒ b = 3a - 6

⇒ - 2a - 6 = 3a - 6 - - - [ From ( 1 ) ]

⇒ - 2a - 3a = - 6 + 6

⇒ - 5a = 0

⇒ a = 0

By substituting a = 0 in equation ( 1 ), we get,

b = - 2a - 6 - - - ( 1 )

⇒ b = - 2 * 0 - 6

⇒ b = 0 - 6

⇒ b = - 6

∴ The values of a and b are ( a, b ) = ( 0, - 6 ).

Answer 2

Answer:

${ \large{ \boxed { \pink { \underline{ \bf \: Option \: D }}}}}$

Step-by-step explanation:

Given :-

Quadratic polynomial = x² + (a + 1)x + b

Zeroes = 2 and -3

To Find :-

Value of a and b

Solution :-

On comparing the equation with ax² + bx + c. We get

a = 1 , b = a + 1 , c = b

We know that

${ \boxed{ \underline{ \sf{ \alpha + \beta = \dfrac{ - b}{a} }}}}$

$\sf \mapsto \: 2 + ( - 3) = \dfrac{ - (a + 1)}{1}$

$\sf \mapsto \: 2 - 3 = \dfrac{ - a - 1}{1}$

$\sf \mapsto \: - 1 = - a - 1$

$\sf \mapsto \: - 1 + 1 = a$

$\sf \mapsto \: 0 = a$

Now

${ \boxed{ \underline{ \sf{ \alpha \beta = \dfrac{c}{a} }}}}$

$\sf \mapsto \: 2 \times ( - 3) = \ \dfrac{b}{1}$

$\sf \mapsto - 6 = b$

Hence

Option D is correct