Question

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

Answer 1

Question

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

Answer:

he values are

a = 0 and b = -6

Step-by-step explanation:

Given , the quadratic polynomial

$\purple{ {x}^{2} + (a + 1)x + b}$

• The roots are 2 and -3

Now from the relationship of the co - efficients of x , x² and constant term with the zeroes of the polynomial we have

$sum \: of \: the \: zeroes = - \frac{co \: efficient \: of \: x}{co \: efficient \: of \: x ^{2} }$

$\implies2 + ( - 3) = - \frac{(a + 1)}{1} \\ \implies2 - 3 = - a - 1 \\ \implies - 1 = - a - 1 \\ \implies a + 1 = 1 \\ \implies \pink{ a = 0}$

And again

$product \: of \: the \: roots \: = \frac{constant \: term}{co \: efficient \: of \: {x}^{2} } \\ \implies 2 \times ( - 3) = \frac{b}{1)} \\ \implies - 6 = b \\ \implies \pink{b = - 6}$

Thus , using the values of a and b quadratic polynomial obtain is

x² + x - 6

Answer 2

Given

2 and -3 are the zeroes of the polynomial

$x^{2} +(a+1)x + b = 0$

Substituting x = 2, we get

$4+2a+2+b=0\\\implies 2a+b=-6 --->(i)$

Substituting x = -3, we get

$9-3a-3+b=0\\\implies -3a+b=-6 --->(ii)$

Solving, we get,

$2a+b=-6\\\underline{-3a+b=-6}\\\underline{\underline{5a = 0}}$

$=>\boxed{\boxed{\bold{ a = 0}}}}$

$=> \boxed{\boxed{\bold{ b = -6}}}}$