Question

white a
a quadratic poly
the sum &
product of whose zeroes are 3&-2.​

Answer 1

Answer:

THE QUADRATIC POLYNOMIAL IS

${x}^{2} - 3x - 2$

Step-by-step explanation:

THE FORMULA TO FIND QUADRATIC POLYNOMIAL IS

$k[ {x}^{2} - ( \alpha + \beta )x - ( \alpha \beta )]$

$here \: \\ \alpha + \beta = 3 \\ \alpha \beta = - 2$

$k[ {x}^{2} - (3)x + ( - 2)] \\ k[ {x}^{2} - 3x - 2] \\ when \: there \: is \: no \: \\ denominator \: in \: the \: \\ given \: zeroes \: then \: the \: \\ k \: value \: becomes \: \\ \\ 1$

$therefore \\ 1( {x}^{2} - 3x - 2) \\ {x}^{2} - 3x - 2$

HOPE IT WILL HELP YOU

PLZZ MARK AS BRAINLIST

Answer 2

GIVEN :–

• Sum of roots = 3

• Product of roots = -2

TO FIND :–

• quadratic polynomial = ?

SOLUTION :–

• We know that a quadratic equation in form of sum of roots and Product of roots is –

$\\ \longrightarrow \large{ \boxed{{ \bold{ {x}^{2} - (sum \: \: of \: \: roots)x + product \: \: of \: \: roots = 0}}}}$

• So that –

$\\ \implies { \bold{ {x}^{2} - 3x + ( - 2)=0 }}$

$\\ \implies \large { \boxed { \bold{ {x}^{2} - 3x - 2=0 }}}$

VERIFICATION :–

$\\ \longrightarrow { \bold{ sum \: \: of \: \: roots = - \dfrac{b}{a} }}$

$\\ \implies { \bold{ 3 = - \dfrac{( - 3)}{1} }}$

$\\ \implies { \bold{ 3 = 3 \: \: \: \: \: (verified) }}$

$\\ \longrightarrow { \bold{ product \: \: of \: \: roots = \dfrac{c}{a} }}$

$\\ \implies { \bold{ - 2 = \dfrac{( - 2)}{1} }}$

$\\ \implies { \bold{ - 2 = - 2\: \: \: \: \: (verified) }}$