Question

find zeros of the quadratic polynomial 6x²-3-7x.​

Answer 1

1/2=7/6

Step-by-step explanation:

=6x²-7x-3

=6x²+2x-9x-3

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

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

⇒2x-3=0                                    or                            ⇒3x+1=0

⇒x=3/2                                       or                           ⇒ x= - 1/3

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    

Answer 2

Answer:

6x²−3−7x=6x²−7x−3=0

⇒6x²+2x−9x−3=0

⇒2x(3x+1)−3(3x+1)=0

⇒(2x−3)(3x+1)=0

Zeros = 3/2 , -1/3

α+β= -b/a ⇒ 3/2 - 1/3 = 7/6 = (-7)/6 = -b/a

αβ= c/a ⇒(3/2) (1/3) = -1/2 = -3/6 = c/a

Step-by-step explanation:

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