Question

find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively 1 over 2, minus 3​

Answer 1

Answer:-

$2 {x}^{2} - x - 6$

Solution :-

Let alpha and beta be the zeroes of the quadratic polynomial.

A/Q.

$\alpha + \beta = \dfrac{1}{2}$

$\alpha \beta = - 3$

The quadratic polynomial is represented in the form of :-

★ $\boxed{\sf{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta }}$

Put the given values,

→${x}^{2} - ( \dfrac{ 1}{2} )x + ( - 3) = 0$

Taking L. C. M

→$\dfrac{ 2{x}^{2} - x + ( - 6) }{2} = 0$

→$2 {x}^{2} - x - 6 = 0$

hence, the required quadratic polynomial will be :-

$2 {x}^{2} - x - 6$

Answer 2

Answer:

2x² - x - 6

Step-by-step explanation:

given that,

In a quadratic polinomial,

sum of zeroes and product of zeroes are respectively 1/2, -3

we know that,

when in a quadratic polinomial zeros are given then quadratic equation will be,

x² - sum of zeros × x + product of zeros

here,

sum of zeros = 1/2

product of zeros = -3

putting the values,

quadratic equation

=> x² - (½)x + (-3)

x² - x/2 - 3 = 0

now,

by multiplying the equation by 2

2(x² - x/2 - 3) = 0 × 2

2x² - 2(x/2) - 3(2) = 0

2x² - x - 6 = 0

by

so,

the required quadratic equation is

2x² - x - 6