Question

If 2 and -3 are the zeroes of the following
quadratic polynomial, then find the value of a
and b

x2 + (a + 1) x + b

Answer 1

Answer

The required values are

a = 0

b = -6

$\bf \large \underline{Given : }$

The quadratic polynomial

• x² + (a + 1)x + b
• 2 and -3 are the zeroes of the given polynomial

$\bf \large \underline{To \: Find : }$

• The value of a and b

$\bf \large \underline{Solution : }$

The zeroes of the polynomial x² + (a + 1)x + b are 2 and -3 .

Thus , from the relationship of sum of zeroes and coefficients we have ,

$\sf sum \: of \: the \: zeroes = \dfrac{coefficients \: of \: x}{coefficient\: of \: x^{2}} \\\\ \sf\implies 2 + (-3) = -\dfrac{a + 1}{1} \\\\ \sf\implies -1 = -a - 1 \\\\ \sf\implies -a = -1+1 \\\\ \sf \implies a = 0$

Again, from the relationship of product of the zeroes and coefficients of the polynomial ,

$\sf product \: of \: the \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\sf\implies (2)(-3) = \dfrac{b}{1} \\\\ \sf\implies -6 = b \\\\ \sf\implies b = -6$

Therefore , required value of a and b are 0 and -6 respectively.

Answer 2

Answer:

The value of a and b are 0 and -6

Step-by-step explanation:

${x}^{2} + ax + x + b = 0$

${2}^{2} + 2a + 2 + b = 0$

$6 + 2a + b = 0$

$2a + b = - 6 \: \: \: \: \: \:$

${ - 3}^{2} - 3a - 3 + b = 0$

$6 - 3a + b = 0$

$- 3a + b = - 6$

$a = 0$

$2(0) + b = - 6$

$b = - 6$