If a and b are the roots of x² - x – 12 – 0, and a > b, then the quadratic equation in x whose roots are (2a – 1) and (2b + 1) is

Correct option is D

Given:

Roots of $$x^2 - x - 12 = 0$$​ are a , b 

New roots: (2a  - 1) , (2b + 1)

​Formula Used:

$$\text{Equation: } x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$$​

Solution: 

Roots of $$x^2 - x - 12 = 0$$​;

​$$x^2 -x - 12 = 0 \\ \ \\ x^2 - 4x +3x-12 = 0 \\ \ \\ x(x-4)+3(x-4) = 0 \\ \ \\ (x-4)(x+3) = 0 \\ \ \\ x = 4 , \ -3$$​​

So,  a = 4 , b = -3 

Now,  

Sum of new roots = (2a-1)+(2b + 1)   

$$\text{Sum of new roots: } 7 + (-5) = 2,$$  

Product of new roots ; (2a-1)(2b + 1)   

= $$(7)(-5) = -35$$​​

So, the new quadratic equation; 

​$$x^2 - 2x - 35 = 0$$​​

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