Two students A and B solve an equation of the form \( \Large x^{2}+px+q=0 \). A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and -9. What are the correct roots of the equation?

Let \( \Large \alpha \) and \( \Large \beta \) be the roots of the

quadratic equation \( \Large x^{2} + px + q = 0 \)

Given that, A starts with a wrong value of p and obtains the roots as 2 and 6.

But this time q is correct.

i.e., product of roots

\( \Large = q = \alpha . \beta = 6 \times 12 = 12 \) ...(i)

and B starts with a wrong value of q and gets the roots as 2 and -9.

But this time p is correct.

i.e., sum of roots ._

\( \Large = p = \alpha + \beta = -9 +2 = -7 \) ...(ii)

\( \Large \left( \alpha - \beta \right)^{2} = \left( \alpha + \beta \right)^{2} - 4 \alpha \beta \)

\( \Large \left(-7\right)^{2} - 4.12 = 49 -48 = 1 \)

[from Eqs. (i) and (ii)]

\( \Large \alpha - \beta = 1 \)

Now, from Eqs. (ii) and (iii), we get

\( \Large \alpha = -3 \  and \   \beta = -4 \)

which are correct roots.