Question

if
The zeros of the Quadractic polynomial x^2 +(a+1)x
+b are 2 and 3 then find a and b​

Answer 1

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

Let assume that,

$\rm :\longmapsto\:f(x) = {x}^{2} + (a + 1)x + b$

Given that

$\rm :\longmapsto\:2 \: and \: 3 \: are \: zeroes \: of \: f(x)$

We know that,

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

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

$\rm :\longmapsto\:5 = - a - 1$

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

Also,

We know that

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

$\rm :\longmapsto\:2 \times 3 = \dfrac{b}{1}$

$\bf\implies \:b = 6$

Additional Information :-

Cubical polynomial

$\rm :\longmapsto\: \alpha, \beta \: and \: \gamma \: are \: zeroes \: of \: {ax}^{2} + bx + c, \: \: then$

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

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

$\green{ \boxed{ \bf \: \alpha \beta \gamma = \frac{d}{a}}}$

2. Useful Identities

$\green{ \boxed{ \bf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }}$

$\green{ \boxed{ \bf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta( \alpha + \beta ) }}$

$\green{ \boxed{ \bf \: {( \alpha - \beta )}^{2} = {( \alpha + \beta) }^{2} - 4 \alpha \beta }}$