Question

Find the zeroes of the polynomial 3x2 + 4x – 4 and verify the relationship between the zeroes and the coefficients.

Answer 1

a = 3

b = 4

c = – 4

3x² + 4x – 4 = 0

3x² + 6x – 2x – 4 = 0

3x ( x + 2 ) – 2 ( x + 2 ) = 0

( x + 2 ) ( 3x – 2 ) = 0

Hence first zero, α = – 2

Second zero, β = 2/3

Sum of the zeroes = α + β

= – 2 + 2/3

= ( – 6 + 2 ) / 3

= – 4/3

Also, sum of zeroes = – b / a

= – 4 / 3

Product of zerores = α × β

= – 2 ( 2 / 3 )

= – 4 / 3

Also, Product of zeroes = c / a

= – 4 / 3

Hence verified

Answer 2

Answer:

Step-by-step explanation:

3x² + 4x – 4

Splitting the middle term, we get,

3x² + 6x – 2x – 4

Taking the common factors out, we get,

3x(x+2) -2(x+2) On grouping, we get,

(x+2)(3x-2)

So, the zeroes are,

x+2=0 ⇒ x= -2

3x-2=0⇒ 3x=2⇒x=2/3

Therefore, zeroes are (2/3) and -2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

– 2 + (2/3) = – (4)/3 =

– 4/3 = – 4/3

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

Product of the zeroes = (- 2) (2/3) = – 4/3