Question

6. Find the quadratic polynomial whose zeros are 1/2and 3/2​

Answer 1

Answer:

k(x^2-2x+3/4)

Step-by-step explanation:

the quadratic polynomial is =k(x^2-(sum of zeros)x+(product of zeros) ,where k is a non zero constant

=k(x^2-2x+3/4)

Answer 2

step by step explanation

given: zeroes of polynomial are 1/2 and 3/2

methods to solve the problem

1. method:

since zeroes of polynomial are given ,

x = 1/2 or x = 3/2

hence we can write the above equation as

x -1/2 = 0 or x -3/2 = 0

so we will get the polynomial we need by multiplying both the factors which are

(x-1/2)(x-3/2) = 0

$x {}^{2} - \frac{3}{2} x - \frac{1}{2} x + \frac{3}{4} = 0$

multiply equation by 4 on both sides

$4x {}^{2} - 2x + 3 = 0$

So the required polynomial is

\boxed{ 2x^2 -4x + 3}

2.method

if alpha and beta are the roots of quadratic equation f(x)

then the polynomial f(X) is given by

\boxed{ f(X)= k[ x^2 -{alpha+ beta}x+ alpha× beta }

f(X) = k[ x²- sum of the zeroes + product of the zeroes ]

sum of zeroes = 3/2 +1/2 = 4/2

product of zeroes = (3/2)× (1/2) = 3/4

So the required polynomial is

\boxed{ k( x^2 -(4/2)x + 3/4}