© examsiri.com
Question : 74 of 200
Marks:
+1,
-0
Solution:
I. 2y2 + 3y – 5 = 0
2y2 + 5y – 2y – 5 = 0
y (2y + 5) –1 (2y + 5) = 0
(2y + 5) (y – 1) = 0
y =
−,1 II.
x2 – 3x = 2x – 6
x2 – 5x + 6 = 0
x2 – 3x – 2x + 6 = 0
x (x – 3) –2 (x – 3) = 0
(x – 3) (x – 2) = 0
x = 3 , 2
Hence, x > y
© examsiri.com
Go to Question: