Question

If α and β are the zeros of the quadratic polynomial f(x) = 6x²+x-2, then the value of

1. α²+β²
2. 1/α+1/β​

Can Someone Answer This Quick Thank You<3

Answer 1

`1) ~\alpha^2+\beta^2 = (25)/(36)\\\\\\2)~\frac 1 {\alpha} + \frac 1{\beta} = \frac 12`

Explanation:

`\text{Given that,}\\\\f(x) = 6x^2 +x -2~ \text{and the roots are}~ \alpha, \beta\\\\\text{Now,}\\\\\alpha + \beta = -(b)/(a) = -(1)/(6)~~~~~;[\text{Compare with the standard form}~ ax^2 +b x + c = 0]\\\\\alpha \beta = \frac ca = -\frac2 6 = - \frac 13\\\\\\\textbf{1)}\\\\\alpha^2 +\beta^2\\\\\\=\left(\alpha +\beta \right)^2 - 2 \alpha \beta \\\\\\=\left( -\frac 16 \right)^2 -2 \left(- \frac 13 \right)\\\\\\=(1)/(36)+\frac 23\\\\\\=(25)/(36)`

`\textbf{2)}\\\\\frac 1{\alpha} + \frac 1{\beta} \\\\\\=(\alpha + \beta)/(\alpha \beta )\\\\\\=(-\tfrac16)/(-\tfrac 13)\\\\\\=(1)/(6) * 3\\\\\\=(1)/(2)`

Answer 2

Given function:

`f(x)=6x^2+x-2`

To find the zeros of the function, set the function to zero and factor:

`\implies 6x^2+x-2=0`

`\implies 6x^2+4x-3x-2=0`

`\implies 2x(3x+2)-1(3x+2)=0`

`\implies (2x-1)(3x+2)=0`

Therefore, the zeros are:

`\implies (2x-1)=0 \implies x=(1)/(2)`

`\implies (3x+2)=0 \implies x=-(2)/(3)`

If α and β are the zeros of the function:

• 
  `\textsf{Let } \alpha=(1)/(2)`

• 
  `\textsf{Let } \beta=-(2)/(3)`

Question 1

`\begin{aligned}\implies \alpha^2+\beta^2 & =\left((1)/(2)\right)^2+\left(-(2)/(3)\right)^2\\\\& = (1)/(4)+(4)/(9)\\\\& = (9)/(36)+(16)/(36)\\\\& = (25)/(36)\end{aligned}`

Question 2

`\begin{aligned}\implies (1)/(\alpha)+(1)/(\beta) & = (1)/((1)/(2))+(1)/(-(2)/(3))\\\\& = 1 * (2)/(1)+1 * -(3)/(2)\\\\& = 2 - (3)/(2)\\\\& = (1)/(2)\end{aligned}`