Question

A quadratic polynomial whose zeroes are 2 and - 3/2 is​

Answer 1

Step-by-step explanation:

For finding quadratic polynomial we need sum of the zeros and product of the zeros , so we will find that first and then put that in the eq•K[x²-(sum)x+product].

If the polynomial is in fraction we have to solve that

Answer 2

Answer:

The quadratic polynomial whose roots are 2 and - 3/2  is   $\frac{1}{2}$ [2x²-x+6]

Step-by-step explanation:

Given,

Zeroes of a polynomial are 2 and $\frac{-3}{2}$

To find,

The equation of the polynomial

Recall the concept

The quadratic polynomial whose zeroes are α, β is given by

x² - (α +β)x +αβ

Solution:

Since the zeroes of the quadratic polynomial are 2 and $\frac{-3}{2}$  we have,

α = 2 and  β =  $\frac{-3}{2}$

Then, sum of roots =  α +β = 2+( $\frac{-3}{2}$) = $\frac{4-3}{2}$ = $\frac{1}{2}$

Product of the roots =  α× β = 2( $\frac{-3}{2}$) = -3

Hence the required polynomial = x² - (α +β)x +αβ

= x² - $\frac{1}{2}$x +-3

=  $\frac{1}{2}$ [2x² - x+6]

∴ The quadratic polynomial whose roots are 2 and - 3/2  is   $\frac{1}{2}$ [2x²-x+6]

SPJ3