Question

give the polynomila of degree 2 with sum and product of its zeros
$\frac{ - 1}{2} and - 3respectively$

Answer 1

Step-by-step explanation:

given,

α+β = -1/2.

αβ = -3 .

quadratic formula is

x^2-x(α+β)+αβ=0.

x^2-x(-1/2)+(-3)=0.

x^2+1/2x-3=0.

multiply both the sides with 2.

2[x^2+1/2x-3]=2[0].

2x^2+x-6=0.

$therefore \:2x^2+x-6=0 \:is\: the\:equation\: with \:sum \:and\:product \:of \:zeroes\:as \: \frac{ - 1}{2} and - 3respectively$

Answer 2

Answer:

$\large{\underline{\boxed{\sf 2x^{2} + x - 6}}}$

Step-by-step explanation:

Let the zeroes of the quadratic polynomial be α and β.

Given that :

Sum of zeroes = $\bf \dfrac{-1}{2}$

⇒ α + β = $\dfrac{-1}{2}$

Product of zeroes = (-3)

⇒ αβ = (-3)

The quadratic polynomial is given by -

∴ x² - (α + β)x + αβ = 0

⇒ x² - ( $\dfrac{-1}{2}$)x + (-3) = 0

⇒ x² + $\dfrac{1}{2}$ x - 3 = 0

⇒ $\sf \dfrac{2x^{2}+x-6}{2}$ = 0

⇒ 2x² + x - 6 = 0

Hence, the required quadratic polynomial is 2x² + x - 6.