AB is parallel to DC in a trapezium ABCD. It is given that AB > DC and the diagonals AC and BD intersect at O. If AO=3x-15, OB = x +9, OC =x-5 and OD = 5, and x has two values x1 and x2, then the value of (x12 + x2 2 ) is:
Given the trapezium with AB || DC and the diagonals intersecting at O: OD/OB =OC/OA 3x-15/x+9 =x-5/ 5 (3x-15)5 =(x+9) ×(x-5) 15x-75 =x² + 4x – 45 x² -11x + 30 = Solve the quadratic equation: x1 = -6, x2 =-5 x1 = 6, X2 =5 Now- (x12 + x2 2 ) = 36 + 25 = 61
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