Question

FIND THE ZEROS OF THE POLIYNOMIAL F(X)=6x^2+x-12 and verify the relationship

Answer 1

F(x) = 6x^2 + x - 12

6x^2 + x - 12 = 0

6x^2 + 9x- 8x - 12 = 0

3x(2x+3) - 4( 2x-3) = 0

(2x+3) ( 3x-4) = 0

2x+3 = 0

x = -3/2

3x - 4 = 0

x = 4/3

the zeroes are 4/3 and -3/2

Product of zeroes

αβ = c/a

4/3 x -3/2 = -12/6

-12/6 = -12/6

Sum of zeroes

α + β = -b/a

4/3 - 3/2 = -1/6

-1/6 = -1/6

Verified

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Answer 2

this is the polynomial in the problem

$F(x) = 6x^2+x-12$

now we need to find the zeros of F(x), there are many ways but I'll go with factoring

$F(x)\\\\=6x^2+x-12\\\\=6x^2+9x-8x-12\\\\=3x(2x+3)-4(2x+3)\\\\=(2x+3)(3x-4)$

therefore, F(x) = (2x + 3) (3x - 4)

but if x is a zero of F(x), F(x) = 0

so,

$F(x)=0\\\\=>(2x+3)(3x-4)=0$

that means either (2x+3) = 0 or (3x-4) = 0

or, x = -3/2 or x = 4/3

let's name them α = -3/2 and β = 4/3

now, the question says to verfy some relationship ... but it is not clear which relation, so I assume you meant relationship of the coefficients of the polynomial and it's zeros

so let's find the coefficients

a = 6            (coefficient of x^2)

b = 1            (coefficient of x^1)

c = -12            (coefficient of x^0, or the constant)

there are two relations between the coefficients and the zeros (for quadratic polynomials),

(i)  α + β = -b / a

(ii) α * β = c / a

so let's verify them,

(i) LHS

               $\alpha + \beta\\\\=-3/2+4/3\\\\=\frac{-9+8}{6}\\\\=\frac{-1}{6}$

  RHS

               $-b/a\\\\=-1/6$

LHS = RHS, therefore (i) is verified

in a similar fashion, (ii) can be verified

(note: if you need to verify some other relation, just put the appropriate values in the variables than evaluate the correct equations ....)

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