Question

find the quadratic polynomial whose sum and the product of whose zeros are root 2 and -3/2 respectively​

Answer 1

Answer:

=x(x-3√2)+2√2(x-3√2)

=(x-3√2)(x+2√2).

∴f(x)=0⇒(x-3√2)(x+2√2)=0.

Answer 2

Answer:

Standard quadratic polynomial

= 2x² - 4x -3 = 0

Step-by-step explanation:

Given:

Sum of zeroes = 2

Product of zeroes = -3/2

Now, a quadratic polynomial in form of zeroes has given format:

K(x² - (sum of zeroes)x + (product of zeroes) = 0

where k is a real number which can be given any real value for getting a multiple of polynomial which is available in option.

Lets take k = 1 and we already know sum and product of zeroes,

∴1(x² -2x + (-3/2)) = 0

= x² - 2x -3/2 = 0

So, this is one form which is obtained when k = 1

Your answer could also be in standard form.

So, For that we have to k = 2:

∴2(x² -2x -3/2) = 0

= 2x² - 4x -3 = 0

If it still doesn't match your options, then you can use different value of k accordingly.

Hope, you got it:-))

Please mark it as brainiest!!