CUET Preparation Today
If 4x2 + 3y2 = 67, xy = 4 (x, y > 0) find x3 + y2. |
64 65 56 76 |
65 |
By Putting x = 4 and y = 1 These values are satisfying the equations: ⇒ 4x2 + 3y2 = 67 ⇒ 4(4)2 + 3(1)2 = 67 ⇒ 67 = 67 Put these values and find: ⇒ x3 + y2 = (4)3 + (1)2= 64 + 1 = 65 |
