Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4,3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (4,3). What would be the exact root of the original quadratic equation ?	Choose the correct choice from the following answers:
A) (6, 1) B) (-3,-4) C) ( 4, 3) D) (-4, -3) E) None of these
Explanation :   Equation of quadratic equation is   ax2 + bx + c = 0 .   x2 + bx + c = 0 . First roots = (4, 3)   Sum of the roots -b / a = -7 .   =&gt; b = 7.   Product of the roots = c / a = 12.   =&gt; c =12.   ∴ Equation formed x2 – 7b + 12 = 0 ............ (i)   Another boy gets the wrong roots (2, 3).   ∴ Sum of the roots = -b / a = -5.   =&gt; b = 5.   Products of the roots = c / a = 6.   =&gt; c = 6.   Equation formed x2 – 5b + 6 = 0 ............ (ii)   x2 + b′x + c1 = 0.   b′ = 2 + 3 .   ∴ c = 6   Hence, x2 – 7x + 6 = 0.   =&gt; x2 – 6x – x + 6 = 0.   =&gt; x(x – 6) – 1(x – 6) = 0.   =&gt; (x – 6)(x – 1) = 0.   ∴ x = 6, 1.   Hence, the actual roots = (6, 1).   (6, 1)nswer : (6, 1)
<result>A</result>
Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4,3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (4,3). What would be the exact root of the original quadratic equation ?