Question

Find the zeros of the polynomial 6x2 - 3 - 7x and verify the relationship between the zeros and the coefficients of the polynomial.

Answer 1

given,
6x²-7x - 3
to find the zeroes of the polynomial,
6x²-7x - 3 = 0
6x²-9x +2x -3 = 0
3x(2x-3) + 1(2x -3) = 0
(3x+1) (2x-3) = 0
x = -1/3 or 3/2
sum of the zeroes = 3/2 - 1/3 = 7/6 = - ( co-efficient of x) / co-efficient of x²
product of the zeroes = -1/3 × 3/2 = -1/2  = constant / co-efficient of x²

Answer 2

Answer:

Question :- 6x²-7x-3

=6x²+2x-9x-3

=2x(3x+1)-3(3x+1)

=(2x-3)(3x+1) = Zeroes !!

⇒2x-3=0                          

         or                          

 ⇒3x+1=0

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

  or                           

⇒ x= - 1/3 ------(2)

Verification :-

α=3/2 ,β= - 1/3

✌α+β= -b/a

⇒3/2+(-1/3)= - (-3)/6

⇒3/2-1/3=1/2

⇒7/6 =1/2

 ✌ αβ=c/a

 ⇒ 3/2(-1/3)= -7/6

⇒-1/2= -7/6

 ⇒1/2=7/6      

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