Question

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

a ) a = 0, b = -6
b ) a = -1, b = 6
c ) a = -2, b = -6​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:2 \: and \: - 3 \: are \: zeroes \: of \: {x}^{2} + (a + 1)x + b$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\:2 + ( - 3) = - \dfrac{(a + 1)}{1}$

$\rm :\longmapsto\: - 1 = - a - 1$

$\rm :\longmapsto\: - 1 + 1= - a$

$\rm :\longmapsto\: 0= - a$

$\bf\implies \:a = 0$

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\:(2) \times ( - 3) = \dfrac{b}{1}$

$\bf\implies \:b = - 6$

• So, Option (a) is correct.

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$