Question

find the zeroes of quadratic polynomial 6x2 - x - 12 and verify the relationship between the zeroes and coefficients​

Answer 1

$\huge \bold{ \underline{ \underline{{Question}}}}$

Find the zeroes of quadratic polynomial 6x2 - x - 12 and verify the relationship between the zeroes and coefficients

$\huge \bold{ \underline{ \underline{ \red{Answer}}}}$

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

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

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

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

so,

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

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 $\tt \: x^2$)

b = 1 (coefficient of $\tt \: x^1$)

c = -12 (coefficient of $\tt \: x^0$, or the constant)

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

• $\large \bold \red{α + β = \frac{-b}{a}}$
• $\large {\bold {\red{α * β = \frac{ c }{ a}}}}$

so let's verify them,

(i) LHS

$\tt =-3/2+4/3\\\\=\frac{-9+8}{6}\\\\=\frac{-1}{6} \tt=−3/2+4/3 \\ \\ \tt== 6−9+8\\ \\ \tt== 6−1 \\ \\ </p><p>\tt=RHS \\ \begin{gathered}-b/a\\\\=-1/6\end{gathered} \\ \\\tt= −b/a \\ \\ \tt=−1/6$

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 ....)

$\huge\bold{ \underline{ \underline{ \blue{@WildCat7083}}}}$