Question

find a quadratic polynomial if it's zeroes are 2 and -1/3 ​

Answer 1

hey

here is ur answer

(x-2)and (x+1/3)

multiply both of them

x-2(x+1/3)

=x²-2x+(x-2)/3

=(3x²-6x+x-2)/3=3x²-5x-2

as there should not be negative power

we can multiply the equation with 3

and the quadratic equation becomes

3x²-5x-2

hope the answer helps u ^_^

=

Answer 2

Solution :

Let the quadratic polynomial be ax²+bx+c and a≠0 and it's zeroes be $\alpha \: and \: \beta}$

Here,

$\alpha = 2 \: and \: \beta = \frac{ - 1}{3}$

Firstly , take the sum of the zeroes

$( \alpha + \beta ) = 2 + ( \frac{ - 1}{3})$

$\frac{5}{3}$

Now take the product of the zeroes

$\alpha \beta = 2( \frac{ - 1}{3} )}$

$\frac{ - 2}{3}$

Note :

The quadratic polynomial ax²+bx+c is k[x²-(a+B)x+aB)where k is constant. Here a is alpha and b is beta

So,

k[x²-5/3x-2/3]

This is derived from the above equation

When k is taken as 3 the quadratic polynomial will be 3x²-5x-2