Math, asked by sridevisridhar200671, 2 months ago

find the zeroes of thefollowing polynomials and verify the releation between zeros and coefficient (a)4x^2 - 3x - 1 (b) 3x^2+4x-4 (c) 5t^2+12+7 (d)t^3 - 2t^2 - 15t (e) 4x^2 +5rootx - 3 (f)7y^2 - 11/3 y-2/3 (g) 2x^2 + 7/2 x +3/4

Answers

Answered by guptavishrut
0

Answer:

Step-by-step explanation:

(a) 4x² – 3x – 1

Splitting the middle term, we get,

4x²-4x+1x-1 Taking the common factors out, we get,

4x(x-1) +1(x-1)

On grouping, we get,

(4x+1)(x-1)

So, the zeroes are, 4x+1= 0⇒ 4x=-1 ⇒x= (-1/4)

(x-1) = 0 ⇒ x=1

Therefore, zeroes are (-1/4) and 1

Verification:

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

α + β = – b/a

1 – 1/4 = – (- 3)/4 = ¾

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

1(- 1/4) = – ¼

– 1/4 = – 1/4

(b) 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

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