Hardik and Shivang attempted to solve a quadratic equation Hardik made a mistake in writing down the constant term and ended up in roots \((4,3)\) Shivang made a mistake in writting down coefficient of \(x\) to get roots \((3,2)\) The correct roots of equation are (a) \(-4,3\) (b) 6,1 (c) 4,3 (d) \(-6,-1\)

Given, Hardik made a mistake in writing the constant term and ended up with roots (4,3). The correct equation has the form \(ax^2+bx+c=0\), and Hardik's incorrect equation has the form \(ax^2+bx+d=0\) with roots 4 and 3. Thus, Hardik's incorrect equation can be written as: \[a(x-4)(x-3)=0\] Shivang made a mistake in writing the coefficient of x and got roots (3,2). The correct equation has the form \(ax^2+bx+c=0\), and Shivang's incorrect equation has the form \(ax^2+kx+c=0\) with roots 3 and 2. Thus, Shivang's incorrect equation can be written as: \[a(x-3)(x-2)=0\]

To find the correct constant term and coefficient of x, let's expand both equations. Hardik's incorrect equation: \[a(x-4)(x-3)=ax^2-7ax+12a=0\] Shivang's incorrect equation: \[a(x-3)(x-2)=ax^2-5ax+6a=0\] Comparing the coefficients, we can see that the coefficients of \(x^2\) are equal in both equations, so the correct equation will also have the coefficient of \(x^2\) as 'a.' Now, let's find the correct coefficient of x. Notice that the middle terms of the two equations differ by 2a (-7a and -5a). So, the correct coefficient of x must be either -5a (if Hardik made a mistake) or -7a (if Shivang made a mistake). Similarly, the constant term differs by 6a (12a and 6a). So, the correct constant term must be either 6a (if Hardik made a mistake) or 12a (if Shivang made a mistake).

Now, let's analyze each option to determine which one is correct. (a) \(-4,3\) If this is the correct option, then the correct equation can be represented as: \[a(x+4)(x-3)=ax^2+ax-12a\] Here, we have a correct coefficient of x as -7a, but not the correct constant term as 12a. Therefore, this option is not correct. (b) 6,1 If this is the correct option, then the correct equation can be represented as: \[a(x-6)(x-1)=ax^2-7ax+6a\] Here, both coefficients of x and the constant term are correct. Therefore, this is the correct option. (c) 4,3 We don't need to check this option, as we have already found the correct answer. (d) \(-6,-1\) We don't need to check this option, as we have already found the correct answer. So, the correct roots of the equation are (Option b) 6 and 1.