Question

Of a and ß are the zeroes of polynomial 3x^2+2x,then find 1/a^2+1/b^2

Answer 1

Answer:

– 14/9

Note:

★ The possible values of variable for which the polynomial becomes zero are called its zeros.

★ In order to find the zeros of the polynomial , equate it to zero.

★ A quadratic polynomial can have atmost two zeros.

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

• Sum of zeros , (α+ß) = -b/a

• Product of zeros , (αß) = c/a

Solution:

Here,

The given quadratic polynomial is :

3x² + 2x + 3 .

Clearly ,

a = 3

b = 2

c = 3

Now,

=> Sum of zeros = -b/a

=> α + ß = -2/3

Also,

=> Product of zeros = c/a

=> αß = 3/3 = 1

Now,

1/α² + 1/ß² = (ß² + α²)/α²ß²

= [ (α + ß)² - 2αß ] / ( αß )²

= [ (-2/3)² - 2 ] / ( 1 )²

= [ 4/9 - 2 ] / 1

= 4/9 - 2

= (4 - 18)/9

= - 14/9

Hence,

The required value of 1/α² + 1/ß² is ;

– 14/9

Answer 2

Answer: 1/α^2 + 1/β^2= -14/9

Step-by-step explanation:

3x^2 + 2x + 3

a= 3, b= 2, c= 3

Let α and β be zeroes then

Sum of zeroes= α + β = -b/a= -2/3

Product of zeroes= α × β = c/a= 3/3= 1

(α + β) ^2= α^2 + β^2 + 2αβ

α^2 + β^2 = (α + β)^2 - 2αβ=(-2/3)^2 - 2 × 1

= (4/9) -2= -14/9

So, 1/α^2 + 1/β^2 = (α^2 + β^2) /α^2β^2

= [-14/9]/ 1^2= -14/9