So, the distance ‘x’ of the pole from gate B satisfies the equation
x 2 + 7x – 60 = 0 So, it would be possible to place the pole if this equation has real roots. To see if this is so or not, let us consider its discriminant. The discriminant is b 2 – 4ac = 72 – 4 × 1 × (– 60) = 289 > 0. So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park. Solving the quadratic equation x 2 + 7x – 60 = 0, by the quadratic formula, we get 7 289 7 17 2 2 -± -± x = = Therefore, x = 5 or – 12. Since x is the distance between the pole and the gate B, it must be positive. Therefore, x = – 12 will have to be ignored. So, x = 5. Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A