If a and b are the roots of  $$x^2+x-2=0$$​, then the quadratic equation in x whose roots are $$\frac{1}{a}+\frac{1}{b}$$​ and ab is:

Correct option is B

Given:

The quadratic equation is:
​$$x^2 + x - 2 = 0$$​​

Formula Used:
Here, the roots of the equation are a and b .
Using the properties of roots of a quadratic equation.
The sum of the roots a + b is given by:
​$$a + b = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -1$$​​
The product of the roots $$a \cdot b$$​ is given by:
​$$a \cdot b = \frac{\text{constant term}}{\text{coefficient of } x^2} = -2$$​​
Solution:  

$$x^2 + x - 2 = 0$$   

Sum  of the roots  a + b =  $$\frac{-1 }{1} = -1$$ 

Product of the roots ab = $$1 \times -2 = - 2$$ 

New equation roots  

​$$\frac{1}{a} \quad \text+ \quad \frac{1}{b} \ and \ ab$$   

Then the sum of the roots of new equation    = $$\frac{1}{a} \quad \text+ \quad \frac{1}{b} \ + \ ab$$    =   $$\frac{a+b}{ab} + ab$$ 

= $$\frac{-1 } {-2} + -2$$   = $$\frac{-3}{2}$$​​                                                                                                          

Product of the roots =  $$(\frac{1}{a} \quad \text+ \quad \frac{1}{b}) \times \ ab \\$$   

= a  + b = -1 

Then the equation whose roots are   $$\frac{1}{a} \quad \text+ \quad \frac{1}{b} \ and \ ab$$   

= $$x^2 - (sum of roots)x + product \ of \ roots = 0$$ 

= $$x^2 - (\frac{-3}{2})x + (-1) = 0 \\\ \\2x^2 + 3x -2 = 0$$​​