Question

If zeros of quadratic polinomiyals is X2(a+1)a+b are2and-3 ,than

Answer 1

Step-by-step explanation:

${\large{\rm{\underline{\orange{Appropriate \: Question :}}}}}$

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

$\\ {\large{\rm{\underline{\orange{Solution :}}}}}$

As given,

❏ x² + (a + 1) x + b is the quadratic polynomial

❏ 2 and -3 are the zeros of quadratic polynomial

Therefore,

${\bold{\rm{: \: ⟹ \: \: \: \: \: \: \: \: \: \: 2 \: + \: (-3) \: = \: \frac{-(a \: + \: 1)}{1}}}}$

${\bold{\rm{: \: ⟹ \: \: \: \: \: \: \: \: \: \: \frac{(a \: + \: 1)}{1} \: = \: 1}}}$

${\bold{\rm{: \: ⟹ \: \: \: \: \: \: \: \: \: \: a \: + \: 1 \: = \: 1}}}$

${\bold{: \: ⟹ \: \: \: \: \: \: \: \: \: \: }}{\bold{\underline{\boxed{\rm{\red{a \: = \: 0}}}}}}$

Let's find the value of b,

${\bold{\rm{: \: ⟹ \: \: \: \: \: \: \: \: \: \: 2 \: × \: (-3) \: = \: b}}}$

${\bold{: ⟹ \: \: \: \: \: \: \: \: \: \: }}{\bold{\underline{\boxed{\rm{\red{b \: = \: -6}}}}}}$

Answer 2

Appropriate Question :-

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

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

Given that

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

We know,

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

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

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

Also,

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

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

$\bf\implies \: - 1 = - (a + 1)$

$\bf\implies \: 1 = a + 1$

$\bf\implies \:a \: = \: 0$

$\begin{gathered}\begin{gathered}\bf\: \bf :\longmapsto\:Hence-\begin{cases} &\sf{a \: = \: 0} \\ &\sf{b \: = \: - \: 6} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

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

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

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

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