Question

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

Answer 1

$\small{ \sf{Given: \: \: 2 \: \: and \: \: -3 \: \: are \: \: the \: \: zeroes \: \: of \: \: the \: \: }}$

$\small{ \sf{quadratic \: \: polynomial \: \: {x}^{2} + (a + 1)x + b. }}$

$\small{\sf{ To \: \: Find: \: \: The \: \: values \: \: of \: \: a \: \: and \: \: b.}}$

$\small{\sf{ In \: \: quadratic \: \: polynomial: \: \: {x}^{2} + (a + 1)x + b}}$

$\small{\sf{ Putting \: \: the \: \: value \: \: 2.}}$

$\tt{ \mapsto \: \: \: {{2}^{2} + (a + 1)2 + b = 0}}$

$\tt{ \mapsto \: \: \: {4 + 2a + 2 + b = 0}}$

$\tt{ \mapsto \: \: \: {6 + 2a + b = 0}}$

$\tt{ \mapsto \: \: \: {b = - 2a - 6}}$

$\small{\sf{ Putting \: \: the \: \: value \: \: - 3.}}$

$\tt{ \mapsto \: \: \: {{(- 3)}^{2} + (a + 1)(-3) + b = 0}}$

$\tt{ \mapsto \: \: \: {9 - 3a - 3 + b = 0}}$

$\tt{ \mapsto \: \: \: {6 - 3a + b = 0}}$

$\tt{ \mapsto \: \: \: {b = 3a - 6}}$

$\small{\sf{ Comparing \: \: equation \: \: (i) \: \: and \: \: (ii).}}$

$\tt{ \mapsto \: \: \: {- 2a - 6 = 3a - 6}}$

$\tt{ \mapsto \: \: \: {- 2a - 3a = - 6 + 6}}$

$\tt{ \mapsto \: \: \: {- 5a = 0}}$

$\mapsto \: \: \: {\boxed{ \tt{a = 0}}}$

$\small{\sf{ Putting \: \: the \: \: value \: \: in \: \: equation \: \: (i).}}$

$\tt{ \mapsto \: \: \: {b = - 2(0) - 6}}$

$\tt{ \mapsto \: \: \: {b = 0 - 6}}$

$\mapsto \: \: \: {\boxed{ \tt{b = - 6}}}$

$\small{\sf{ Hence,}}$

$\small{\sf{ The \: \: values \: \: of \: \: a \: \: and \: \: b \: \: are \: \: 0 \: \: and \: \: - 6 \: \: respectively.}}$

Answer 2

Given :-

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

To Find :-

Value of a and b

Solution :-

α + β = -b/a

It may be also written as -(α + β)

-(α + β) = b/a

-[2 + (-3)] = a + 1/1

-[2 - 3] = a + 1

-[-1] = a + 1

1 = a + 1

a = 1 - 1

a = 0

Now

αβ = c/a

2 × (-3) = b/1

-6 = b

Hence

Value of a is 0 and b is -6