Question

.Find the zeroes of the polynomial 2x2 -9 - 3x and then find the relation between zeroes andS coefficients.

Answer 1

ANSWER:

• Zeros are 3 and -3/2

GIVEN:

• p(x)= 2x²-9-3x

TO VERIFY :

• Relationship between zeros and coefficients.

SOLUTION:

Find zeros:

=> 2x²-3x-9 = 0

=> 2x²-6x+3x-9 = 0

=> (2x²-6x)+(3x-9) = 0

=> 2x(x-3)+3(x-3) = 0

=> (x-3)(2x+3) = 0

Either (x-3) = 0

=> x = 3

Either (2x+3) = 0

=> 2x = -3

=> x = -3/2

Finding sum of zeros :

= 3+(-3/2)

= 6-3/2

= 3/2

Now

=> -(-3)/2 = -(Coefficient of x)/(coefficient of x²)

Finding product of zeros:

= 3(-3/2)

= -9/2

=> -9/2 = (Constant term)/(Coefficient of x²)

NOTE:

=> Sum of zeros (α+β) = -(Coefficient of x)/Coefficient of x²

=> Product of zeros (αβ) = Constant term/ Coefficient of x²

Answer 2

$\huge\mathfrak\green{Answer:}$

Given:

We have been given a polynomial 2x^2 - 9 - 3x.

To Find:

We need ti find the zeroes of this polynomial and also we need to verify the relation between zeroes and coefficients.

Solution:

The given polynomial is 2x^2 - 9 - 3x.

or 2x^2 -3x - 9

We can find the zeroes of this polynomial by the method of splitting the middle term.

We need to find two such numbers whose sum is -3 and product is -18.

The two numbers are -6 and +3.

So we get,

2x^2 - 6x + 3x - 9 = 0

= 2x(x - 3) + 3(x - 3) = 0

= (x - 3)(2x + 3)

Either x - 3 = 0 or 2x + 3 = 0.

When x - 3 = 0

=> x = 3.

When 2x + 3 = 0

=> 2x = -3

=> x = -3/2

Now, the two zeroes of the polynomial are 3 and -3/2.

Inorder to verify the relation between zeroes and coefficients, we have

Sum of zeroes(α + β)

= 3 + (-3/2)

= 3 - 3/2

= (6-3)/2

= 3/2 = -b/a

Product of zeroes(αβ)

= 3 × (-3/2)

= -9/2 = c/a

Hence the relation between zeroes and coefficients of the polynomial is verified.