Question

find a quadratic polynomial whose zeros are -3 and 2​

Answer 1

Given :

• The zeros of the quadratic equation is -3 and 2

$\bullet\sf{\alpha =-3}$

$\\ \bullet\sf{\beta=2}$

To Find :

• The quadratic equation =?

Formula :

$\bullet \sf \: sum \: of \: the \: zeros = \alpha + \beta$

$\\ \bullet \sf \: the \: product \: of \: zeros = \alpha \beta$

The formation of quadratic equation is :

$\red \bigstar \boxed{ \sf{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0}}$

Solution :

$\bullet \sf \: \alpha = - 3$

$\bullet \sf \: \beta = 2$

$\implies \bf \: sum \: of \: zeros = \alpha + \beta$

$\\ \implies \sf \: sum \: of \: zeros = - 3 + 2 \\ \\ \implies \sf \: sum \: of \: zeros = 1$

$\implies \bf \: product \: of \: zeros = \alpha \beta$

$\\ \implies \sf product \: of \: zeros = \alpha \beta \\ \\ \implies \sf \: product \: of \: zeros = - 3 \times 2 \\ \\ \implies \sf \: product \: of \: zeros = - 6$

By using Formation of quadratic equation :

$\implies \sf \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ \\ \implies \sf \: {x}^{2} - ( - 3 + 2)x + ( - 3 \times 2) = 0 \\ \\ \implies \sf \: {x}^{2} - ( - 1)x - 6 = 0 \\ \\ \implies \sf \: {x}^{2} + x - 6 = 0$

The Quadratic equation will be x²+x-6=0 .

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{zeros \: of \: polynomial} \\ &\sf{ - \: 3 \: and \: 2} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find - \begin{cases} &\sf{a \: quadratic \: polynomial}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$: \implies \tt \: Let \: \alpha = - 3 \: and \: \beta = 2$

$: \implies \tt \: Sum \: of \: zeros \: = \alpha + \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \tt \: - 3 \: + \: 2$

$: \implies \tt \: \boxed{ \purple{ \rm \: \alpha + \beta = - 1}}$

★ Now,

$: \implies \tt \: Product \: of \: zeros \: = \alpha \beta = ( - 3) \times (2)$

$: \implies \tt \: \boxed{ \purple{ \rm \: \alpha \beta = - 6}}$

★ We know that, the quadratic polynomial is given by

$\tt \:  f(x) = k \bigg( {x}^{2} - (Sum \: of \: zeros)x + Product \: of \: zeros\bigg)$

★ where, k is non zero real number.

$: \implies \tt \: f(x) = k \bigg( {x}^{2} - ( \alpha + \beta )x + \alpha \beta \bigg)$

$: \implies \tt \: f(x) = k \bigg( {x}^{2} - ( - 1 )x - 6 \bigg)$

$: \implies \tt \: f(x) = k \bigg( {x}^{2} + x - 6 \bigg)$