Question

if alpha and beta are the zeros of polynomial 6x square + X - 2 find the value of Alpha upon beta + beta fun Alpha​

Answer 1

Answer:

Let’s first solve the equation to get our zeroes of the equation i.e. alpha and beta.

6x2+x−2=0

6x2+4x−3x−2=0

2x(3x+2)−(3x+2)=0

which gives us two equations,

2x−1=0=>x=1/2(alpha)

3x+2=0=>x=−2/3(beta)

So, the value of alpha/beta and beta/alpha will be -3/4 and -4/3, adding both of them gives the solution as -25/12.

Answer 2

Given: $\alpha \text { and } \beta$ are the zeroes of polynomial $6x^2+x-2$

To find: Value of  $\dfrac{1}{\alpha} +\dfrac{1}{\beta}$

Step-by-step explanation:

As we know If $ax^2+bx+c$ is a quadratic polynomial then

$\alpha+\beta=\dfrac{-b}{a}$ and $\alpha \beta = \dfrac{c}{a}$

Therefore by comparing  $6x^2+x-2$

we get

$\alpha +\beta = \dfrac{-1}{6} \text { and } \alpha \beta = \dfrac{-2}{6}$

Now to find

$\dfrac{1}{\alpha} +\dfrac{1}{\beta}\\\\=\dfrac{\alpha +\beta}{\alpha\beta} =\dfrac{\dfrac{-1}{6} }{\dfrac{-2}{6}} =\dfrac{1}{2}$

Hence the value of  $\dfrac{1}{\alpha} +\dfrac{1}{\beta} \text { is } \dfrac{1}{2}$