1,380 views

0 votes

Let z=(x/y). Now, c = 16z + 49/z => 16z$^{2}$ – cz + 49 = 0. Since x & y are non zero real number. So, z is also a non zero real number. So, the equation 16z$^{2}$ – cz + 49 = 0 satisfies for some real number z, if b$^{2}$ – 4ac >= 0 => c$^{2}$ – 56$^{2}$ >= 0 => |c| >= 56.

Therefore the absolute value of c is always >= 56. So, value of c can’t be -50.

Ans is (B).

Therefore the absolute value of c is always >= 56. So, value of c can’t be -50.

Ans is (B).